MTH 221: Linear Algebra at The American University of Sharjah

 

1. List of concepts and definitions                     Back to the main page (home)

 

2.  For Examples that Illustrate some of the concepts Click here (Linear Algebra Tool Kit)

 

3. Important RESULTS (THEOREMS)(See the Table Below):

       1 About homogeneous system of equations
        2 About nonsingular (invertible) matrices
         3 About elementary matrices
         4 About symmetric matrices
         5 About triangular matrices
        6 About skew-symmetric matrices
        7 About Determinant of a matrix
         8 About basis, dependent, independent
          9 About Linear Transformations
        10 dim(column space of A ) + dim(Nul(A) ) = number of  columns of A
         11 About orthogonal vectors
        12 About Eigenvalues, Eigenspaces , and diagolization

 

 

  Back to the  MAIN PAGE of  Ayman Badawi

 

 

 

 

 

 

 

 

 

 

 


 

 

 

 

Eiginvalues and eigenspaces result

 

Theorem    If $ A$ is an $ n\times n$ matrix, then
(a)
Det $ (A-\lambda I)$ is a polynomial $ p(\lambda)$ of degree $ n$.
(b)
The eigenvalues of $ A$ are the solutions of $ p(\lambda) = 0$.
(c)
If $ \lambda_0$ is an eigenvalue, any nontrivial solution of $ (A-\lambda_0I)X = 0$ is an eigenvector of $ A$ corresponding to $ \lambda_0$.
(D)
If $ A$ and $ B$ are similar, then the set of eigenvalues of $ A$ is equal to the set of eigenvalues of $ B$. If the similarity transform is $ P$ (that is, if $ A
= P^{-1}BP$) and $ (\lambda_0,X)$ is an eigenpair of $ A$, then $ (\lambda_0,PX)$ is an eigenpair of $ B$.
(E)  If $ A_{n\times n}$ is diagonalizable, then $ A$ has $ n$ linearly independent eigenvectors. Also, in the equation $ A = PDP^{-1}, P$ is a matrix whose columns are eigenvectors, and the diagonal entries of $ D$ are the eigenvalues corresponding column by column to their respecctive eigenvectors.

(F)    

If $ A_{n\times n}$ has $ n$ linearly independent eigenvectors $ X_1,X_2,\ldots,
X_n$ with

 

$\displaystyle AX_1 = \lambda_1X_1 \cdots AX_n = \lambda_nX_n$

 

then $ A$ is diagonalizable and $ A = PDP^{-1}$, where

 

$\displaystyle P(X_1X_2\cdots X_n) D = \left(\begin{matrix}
\lambda_1&&&&{\bf0}\  &&\lambda_2\  &&&\ddots\  &{\bf0}&&&&\lambda_n
\end{matrix}\right)$
 

 Eigenvalues and Eigenvectors

Have you ever heard the words eigenvalue and eigenvector? They are derived from the German word "eigen" which means "proper" or "characteristic." An eigenvalue of a square matrix is a scalar that is usually represented by the Greek letter lambda (pronounced lambda). As you might suspect, an eigenvector is a vector. Moreover, we require that an eigenvector be a non-zero vector, in other words, an eigenvector can not be the zero vector. We will denote an eigenvector by the small letter x. All eigenvalues and eigenvectors satisfy the equation equation for a given square matrix, A.

 

Remark 27 Remember that, in general, the word scalar is not restricted to real numbers. We are only using real numbers as scalars in this book, but eigenvalues are often complex numbers.

 

Definition 9.1 Consider the square matrix A . We say that lambda is an eigenvalue of A if there exists a non-zero vector x such that equation. In this case, x is called an eigenvector (corresponding to lambda ), and the pair (lambda ,x) is called an eigenpair for A.

Let's look at an example of an eigenvalue and eigenvector. If you were asked if eigenvector is an eigenvector corresponding to the eigenvalue eigenvalue for matrix A, you could find out by substituting x, lambda and A into the equation Ax=lambda x.
 

 

 


Therefore, lambda and x are an eigenvalue and an eigenvector, respectively, for A.

Now that you have seen an eigenvalue and an eigenvector, let's talk a little more about them. Why did we require that an eigenvector not be zero? If the eigenvector was zero, the equation Ax=lambda x would yield 0 = 0. Since this equation is always true, it is not an interesting case. Therefore, we define an eigenvector to be a non-zero vector that satisfies Ax=lambda x. However, as we showed in the previous example, an eigenvalue can be zero without causing a problem. We usually say that x is an eigenvector corresponding to the eigenvalue lambda if they satisfy Ax=lambda x Since each eigenvector is associated with an eigenvalue, we often refer to an x and lambda that correspond to one another as an eigenpair. Did you notice that we called x "an" eigenvector rather than "the" eigenvector corresponding to lambda? This is because any non-zero, scalar multiple of an eigenvector is also an eigenvector. If you let c represent a scalar, then we can prove this fact through the following steps.
 

 

A(cx)=cAx=clambda x=lambda (cx)

 

Since any non-zero, scalar multiple of an eigenvector is also an eigenvector, eigenvector and eigenvector are also eigenvectors corresponding to lambda =0 when matrix A

Let's find both of the eigenvalues of the matrix matrix A
 

eigenvalues 6 and 1


Therefore, lambda =6 or lambda =1 We now know our eigenvalues. Remember that all eigenvalues are paired with an eigenvector. Therefore, we can substitute our eigenvalues, one at a time, into the formula (A-lambda I)x=0 and solve to find a corresponding eigenvector.

Let's find an eigenvector corresponding to lambda =6


eigenvector


Notice that this system is underdetermined. Therefore, there are an infinite number of solutions. So, any vector that solves the equation x1 - 2x2 = 0 is an eigenvector corresponding to lambda =6 when matrix A To have a consistent method for finding an eigenvector, let's choose the solution in which x2 = 1. We can use back-substitution to find that x1 - 2(1) = 0 which implies that x1 = 2. This tells us that eigenvector is an eigenvector corresponding to lambda =6 when matrix A This is the same solution that we found when we used the power method to find the dominant eigenpair.

Let's find an eigenvector corresponding to lambda =1
 

 

eigenvector


Notice that this system is underdetermined. This will always be true when we are finding an eigenvector using this method. So, any vector that solves the equation x1 + 3x2 = 0 is an eigenvector corresponding to the eigenvalue lambda =1 when matrix A Again, let's choose the eigenvector in which the last element of x is 1. Therefore, x2 = 1 and x1 + 3(1) = 0, so x1 = -3. This tells us that eigenvector is an eigenvector corresponding to lambda =1 when matrix A Using the characteristic equation and Gaussian elimination, we are able to find all the eigenvalues to the matrix and corresponding eigenvectors.

Let's find the eigenpairs for the matrix matrix A.
 

 

eigenpairs 4 and -4


Therefore, lambda =4 or lambda =-4.

Let's find an eigenvector corresponding to lambda =4
 

 

eigenvector


Since the system is underdetermined, we have an infinite number of solutions. Let's choose the solution in which x2 = 1. We can use back-substitution to find that x1 - 3(1 ) = 0 which implies that x1 = 3. This tells us that eigenvector is an eigenvector corresponding to lambda =4 when matrix A

Let's find an eigenvector corresponding to lambda =-4
 

 

eigenvector


Again, let's choose the eigenvector in which the last element of x is 1. Therefore, x2 = 1 and x1 + 1(1) = 0, so x1 = -1. This tells us that eigenvector is an eigenvector corresponding to lambda =-4 when matrix A Using the characteristic equation and Gaussian elimination, we are able to find both of the eigenvalues to the matrix and corresponding eigenvectors.

We can find eigenpairs for larger systems using this method, but the characteristic equation gets impossible to solve directly when the system gets too large. We could use approximations that get close to solving the characteristic equation, but there are better ways to find eigenpairs that you will study in the future. However, these two methods give you an idea of how to find eigenpairs.

Another matrix for which the power method will not work is the matrix matrix a because the eigenvalues are both the real number 5. The method that we showed you earlier will yield the eigenvector eigenvector to correspond to the eigenvalue lambda =5. Other methods will reveal, and you can check, that eigenvector is also an eigenvector of A corresponding to lambda =5. Notice that these two eigenvectors are not multiples of one another. If the same eigenvalue is repeated p times for a particular matrix, then there can be as many as p different eigenvectors that are not multiples of each other that correspond to that eigenvalue.

We said that eigenvalues are often complex numbers. However, if the matrix A is symmetric, then the eigenvalues will always be real numbers. As you can see from the problems that we worked, eigenvalues can also be real when the matrix is not symmetric, but keep in mind that they are not guaranteed to be real.

Did you know that the determinant of a matrix is related to the eigenvalues of the matrix? The product of the eigenvalues of a square matrix is equal to the determinant of that matrix. Let's look at the two matrices that we have been working with. For matrix A
 

 

product of eigenvalues

 


For matrix A
 

 

product of eigenvalues


You can use this as a check to see that you have the correct eigenvalues and determinant for the matrix A.

Example    Show that

 

$\displaystyle A = \left(\begin{array}{rcc} 0&1&0\  -1&0&0\  0&0&2\end{array}\right)$

 

has only one real eigenvalue. Find the eigenspace corresponding to that eigenvalue.

 

Solution The characteristic polynomial is

 

$\displaystyle \det \left(\begin{array}{rrc} -\lambda&1&0\  -1&-\lambda&0\  0&0&2-\lambda
\end{array}\right) = (\lambda^2 +1)(2-\lambda)$

 

Since $ \lambda^2+1$ is never zero for real $ \lambda, \lambda=2$ is the only real eigenvalue. The eigenvector is found by solving $ A - \lambda I = 0$ with $ \lambda=2$:

 

$\displaystyle \left(\begin{array}{rrc\vert c} -2&1&0&0\  -1&-2&0&0\\
0&0&0&0...
...egin{array}{crc\vert c} 1&-\frac12&0&0\  0&1&0&0\\
0&0&0&0\end{array}\right)$

 

The general solution is

 

$\displaystyle \left(\begin{matrix}0\  0\  k\end{matrix}\right)$

 

and the eigenpair is

 

$\displaystyle \left(2, \left(\begin{matrix}0\  0\  k\end{matrix}\right)\right)$

 

The eigenspace

 

$\displaystyle E_{(2)} = {\text{span }} \left\{\left(\begin{matrix}0\  0\\
1\end{matrix}\right)\right\}$

 

 

 

Example    Solve the eigenproblem for

 

$\displaystyle A = \left(\begin{array}{rrr} -4&-4&-8\  4&6&4\  6&4&10\end{array}\right)$

 

Solution The characteristic polynomial $ p(\lambda)$ is

 

$\displaystyle \det \left(\begin{matrix}-4-\lambda&-4&-8\  4&6-\lambda&4\  6&4&10-\lambda
\end{matrix}\right) = -\lambda^3 + 12\lambda^2 - 44\lambda + 48$

 

To solve $ p(\lambda) = 0$, we first guess at one root of $ p(\lambda)$. First we try integer factors of 48: $ \pm 1$, $ \pm 2$, $ \pm 3$, $ \pm 4$, $ \pm 6$, $ \pm
8$, $ \pm 12$, $ \pm 16$, $ \pm 24$, $ \pm 48$. Since $ p(1) \ne 0$ and $ p(-1)\ne
0$, neither 1 nor $ -1$ is a root. However, $ p(2)=0$, so 2 is a root and $ \lambda-2$ is a factor. Dividing, we have

 

  $\displaystyle   -\lambda^2 + 10\lambda - 24$    
$\displaystyle \lambda-2$ $\displaystyle \sqrt{-\lambda^2 + 12\lambda^2 - 44\lambda + 48}$    
  $\displaystyle    \underline{-\lambda^3+ \phantom{1}2\lambda^2 \phantom{-44\lambda}}$    
  $\displaystyle     \phantom{-\lambda^3+} 10\lambda^2 - 44\lambda$    
  $\displaystyle     \phantom{-\lambda^3+} \underline{10\lambda^2 - 20\lambda \phantom{+ 48}}$    
  $\displaystyle    \phantom{-\lambda^3+ 10\lambda^2} -24\lambda + 48$    
  $\displaystyle     \phantom{-\lambda^3+ 10\lambda^2} \underline{-24\lambda + 48}$    
  $\displaystyle   \phantom{-\lambda^3+ 10\lambda^2 -24\lambda + 4}0$    

 

 

so that

 

$\displaystyle p(\lambda) = (\lambda-2)(-\lambda^2 + 10\lambda - 24)$

 

Now we can factor the quadratic part of $ p(\lambda)$ easily into $ (-\lambda+4)(\lambda-6)$. Therefore

 

$\displaystyle p(\lambda) = (\lambda-2) (-\lambda+4)(\lambda-6)$

 

and the eigenvalues are 2, 4, and 6.

To find the eigenvectors, we substitute $ \lambda=2$ $ \lambda=4$, and $ \lambda=6$ into $ (A-\lambda I)X=0$ and solve. The resulting equations are

 

$\displaystyle \lambda$ $\displaystyle =2\colon$ $\displaystyle \left(\begin{array}{rrr\vert c} -6&-4&-8&0\ 4&4&4&0\ 6&4&8&0\en...
...array}{ccr\vert c} 3&2&4&0\ 0&4&-4&0\ 0&0&0&0\end{array}\right) \Rightarrow X$ $\displaystyle = \left(\begin{array}{r} -2k\ k\ k\end{array}\right)$    
$\displaystyle \lambda$ $\displaystyle =4\colon$ $\displaystyle \left(\begin{array}{rrr\vert c} -8&-4&-8&0\ 4&2&4&0\ 6&4&6&0\en...
...{array}{ccc\vert c} 2&1&2&0\ 0&4&0&0\ 0&0&0&0\end{array}\right) \Rightarrow X$ $\displaystyle = \left(\begin{array}{r} -j\ 0\ j\end{array}\right)$    
$\displaystyle \lambda$ $\displaystyle = 6\colon$ $\displaystyle \left(\begin{array}{rrr\vert c} -10&-4&-8&0\ 4&0&4&0\ 6&4&4&0\e...
...ay}{rrr\vert c} -5&-2&-4&0\ 0&-2&1&0\ 0&0&0&0\end{array}\right) \Rightarrow X$ $\displaystyle = \left(\begin{array}{r} -r\ \frac12 r\ r\end{array}\right)$    

 

 

The eigenpairs are

 

$\displaystyle \left(2, \left(\begin{array}{r} -2k\  k\  k\end{array}\right)\r...
...)
\left(6, \left(\begin{array}{r} -r\  \frac12 r\  r\end{array}\right)\right)$

 

and the eigenspaces are

 

$\displaystyle E_{(2)} = {\text{span }} \left\{\left(\begin{array}{r} -2\  1\ ...
...t{span }} \left\{\left(\begin{array}{r} -2\  1\\
2\end{array}\right)\right\}$

 

 

 

Example    The matrices

 

$\displaystyle A = \left(\begin{array}{crc} 3&-1&0\  0&3&0\  0&0&5\end{array}\...
...{and}} B = \left(\begin{array}{ccr} 3&0&0\  0&3&0\\
0&0&-5\end{array}\right)$

 

both have eigenvalues 3 and $ -5$, with 3 being an eigenvalue of multiplicity 2. Compare the eigenspaces for these two matrices.

 

Solution For $ A$ the general eigenpairs are

 

$\displaystyle \left(3, \left(\begin{matrix}k\  0 \  0\end{matrix}\right)\righ...
...text{and}} \left(-5, \left(\begin{matrix}0\  0\\
r\end{matrix}\right)\right)$

 

so the eigenspaces are

 

$\displaystyle E_{(3)} = {\text{span }} \left\{\left(\begin{matrix}1\  0\\
0\...
...text{span }} \left\{\left(
\begin{matrix}0\  0\  1\end{matrix}\right)\right\}$

 

For $ B$ the general eigenpairs are

 

$\displaystyle \left(3, \left(\begin{matrix}r\  s\  0\end{matrix}\right)\right...
...text{and}} \left(-5, \left(\begin{matrix}0\  0\\
t\end{matrix}\right)\right)$

 

so the eigenspaces for $ B$ are

 

$\displaystyle E_{(3)} =$   span $\displaystyle \left\{\left(\begin{matrix}1\  0\\
0\end{matrix}\right), \left(\begin{matrix}0\  1\\
0\end{matrix}\right)\right\} E_{(-5)} =$   span $\displaystyle \left\{\left(
\begin{matrix}0\  0\  1\end{matrix}\right)\right\}$

 

For $ A, \dim E_{(3)} <$ (multiplicity of $ \lambda=3$), while for $ B$, $ \dim
E_{(3)} =$ (multiplicity of $ \lambda=3$).

 

 

 


 

 

a
Remark 32 The eigenvalue of smallest magnitude of a matrix is the same as the inverse (reciprocal) of the dominant eigenvalue of the inverse of the matrix. Since most applications of eigenvalues need the eigenvalue of smallest magnitude, the inverse matrix is often solved for its dominant eigenvalue. This is why the dominant eigenvalue is so important.

Also, a bridge in Manchester, England collapsed in 1831 because of conflicts between frequencies. However, this time, the natural frequency of the bridge was matched by the frequency caused by soldiers marching in step. Large oscillations occurred and the bridge collapsed. This is why soldiers break cadence when crossing a bridge.

Frequencies are also used in electrical systems. When you tune your radio, you are changing the resonant frequency until it matches the frequency at which your station is broadcasting. Engineers used eigenvalues when they designed your radio.

Frequencies are also vital in music performance. When instruments are tuned, their frequencies are matched. It is the frequency that determines what we hear as music. Although musicians do not study eigenvalues in order to play their instruments better, the study of eigenvalues can explain why certain sounds are pleasant to the ear while others sound "flat" or "sharp." When two people sing in harmony, the frequency of one voice is a constant multiple of the other. That is what makes the sounds pleasant. Eigenvalues can be used to explain many aspects of music from the initial design of the instrument to tuning and harmony during a performance. Even the concert halls are analyzed so that every seat in the theater receives a high quality sound.

Car designers analyze eigenvalues in order to damp out the noise so that the occupants have a quiet ride. Eigenvalue analysis is also used in the design of car stereo systems so that the sounds are directed correctly for the listening pleasure of the passengers and driver. When you see a car that vibrates because of the loud booming music, think of eigenvalues. Eigenvalue analysis can indicate what needs to be changed to reduce the vibration of the car due to the music.

Eigenvalues are not only used to explain natural occurrences, but also to discover new and better designs for the future. Some of the results are quite surprising. If you were asked to build the strongest column that you could to support the weight of a roof using only a specified amount of material, what shape would that column take? Most of us would build a cylinder like most other columns that we have seen. However, Steve Cox of Rice University and Michael Overton of New York University proved, based on the work of J. Keller and I. Tadjbakhsh, that the column would be stronger if it was largest at the top, middle, and bottom. At the points 1/4 of the way from either end, the column could be smaller because the column would not naturally buckle there anyway. A cross-section of this column would look like this:
 

 

cylinder

 

Does that surprise you? This new design was discovered through the study of the eigenvalues of the system involving the column and the weight from above. Note that this column would not be the strongest design if any significant pressure came from the side, but when a column supports a roof, the vast majority of the pressure comes directly from above.

Eigenvalues can also be used to test for cracks or deformities in a solid. Can you imagine if every inch of every beam used in construction had to be tested? The problem is not as time consuming when eigenvalues are used. When a beam is struck, its natural frequencies (eigenvalues) can be heard. If the beam "rings," then it is not flawed. A dull sound will result from a flawed beam because the flaw causes the eigenvalues to change. Sensitive machines can be used to "see" and "hear" eigenvalues more precisely.

Oil companies frequently use eigenvalue analysis to explore land for oil. Oil, dirt, and other substances all give rise to linear systems which have different eigenvalues, so eigenvalue analysis can give a good indication of where oil reserves are located. Oil companies place probes around a site to pick up the waves that result from a huge truck used to vibrate the ground. The waves are changed as they pass through the different substances in the ground. The analysis of these waves directs the oil companies to possible drilling sites.

There are many more uses for eigenvalues, but we only wanted to give you a sampling of their uses. When you study science or engineering in college, you will become quite familiar with eigenvalues and their uses. There are also numerical difficulties that can arise when data from real-world problems are used. 

 

    

 

 

 

 

 

 

 

 IMPORTANT RESULTS:

Fact 1 (about homogeneous system of equations):

A system of linear equation is called homogeneous if the right sides are equal to 0.

Example:

                                      2x + 3y - 4z = 0
                                       x -  y +  z = 0
                                       x -  y      = 0
 

A homogeneous system of equations always has a solution (0,0,...,0). Every homogeneous system has either exactly one solution or infinitely many solutions. If a homogeneous system has more unknowns than equations, then it has infinitely many solutions.

 

Fact 2( about nonsingular (invertible) matrices):

If A is an n by n square matrix, then the following statements are equivalent.

  1. A invertible (nonsingular)
  2. The system Av=b has at least one solution for every column-vector b.
  3. The system Av=b has exactly one solution for every column-vector b (here v is the column-vector of unknowns).
  4. The system Av=0 has only the trivial solution (0,0,0,...0)
  5. The Reduced-echelon-form of  A is the identity matrix  n by n.
  6. A is a product of elementary matrices.

 

 

Fact 3 ( about elementary matrices):

If the elementary matrix E is obtained by performing a row operation on the identity  matrix Im and if A is an m by n matrix, then the product EA is the matrix obtained from A by applying the same row operation.

Every elementary matrix is invertible and the inverse is again an elementary matrix. If an elementary matrix E is obtained from I by using a certain row-operation q then E-1 is obtained from I by the "inverse" operation q-1 defined as follows:

  If q is the adding operation (add x times row j to row i) then q-1 is also an adding operation (add -x times row j to row i).

  If q is the multiplication of a row by x then q-1 is the multiplication of the same row by x-1.

  If q is a swapping operation then q-1=q.

 

Fact 4( about symmetric matrices):

  1. The sum of two symmetric matrices is a symmetric matrix.
  2. If we multiply a symmetric matrix by a scalar, the result will be a symmetric matrix.
  3. If A and B are symmetric matrices then AB+BA is a symmetric matrix
  4. Any power An of a symmetric matrix A (n is any positive integer) is a symmetric matrix.
  5. If A is an invertible symmetric matrix then A-1 is also symmetric.
  6. If the product of two symmetric matrices A and B of the same size is symmetric then AB=BA.
  7. Conversely, if A and B are symmetric matrices of the same size and AB=BA then AB is symmetric

 

Fact 5(about triangular matrices):

 

  1. Every square matrix is a sum of an upper triangular matrix and a lower triangular matrix.
  2. The product of two upper (lower) triangular matrices is an upper (lower) triangular matrix.
  3. The transpose of an upper triangular matrix is a low triangular matrix.

 

Fact 6(about skew-symmetric matrices):

 

  1. If A is invertible and skew-symmetric matrices then the inverse of A is skew-symmetric.
  2. If A and B are skew-symmetric matrices then AT, A+B, AB-BA, and kA are skew-symmetric for every scalar k.
  3. Every square matrix is the sum of a symmetric and a skew-symmetric matrices.

 

Fact 7(about Determinant of a matrix):

 

Let A be an n by n matrix. Then the following conditions hold.

  1. If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number.
  2. If two rows (columns) in A are equal then det(A)=0.
  3. If we add a row (column) of A multiplied by a scalar k to another row (column) of A, then the determinant will not change.
  4. If we swap two rows (columns) in A, the determinant will change its sign.
  5. det(A)=det(AT).
  6. If A has a zero row (column) then det(A)=0.
  7. det(AB)=det(A)det(B) for any B, n by n matrices
  8. Aadj(A) = det(A)I =adj(A)A.
  9.   If A is invertible (nonsingular) then A-1 = (1/det(A)) adj(A)

 

Fact 8(about basis, dependent, independent)

Let V be a vector space. The following properties of bases of V hold:

  1. If S is a basis of a vector space V then every vector in V has exactly one representation as a linear combination of elements of S.
  2. If V has a basis with n elements then
    1. Every set of vectors in V which has more than n elements is linearly dependent.
    2. Every set of vectors with fewer than n elements does not span V.
    3. Every set of  n  independent vectors in  V  form a basis for V
  3. All bases of V have the same number of elements.
  4. If V is an n-dimension space and S is a set of n elements from V. Then S is a basis of V in each of the following cases:
    1. S spans V.
    2. S is linearly independent.
  5. If S is a linearly dependent set in an n-dimensional space V and V=span(S) then by removing some elements of S we can get a basis of V.
  6. If S is a linearly independent subset of V which is not a basis of V then we can get a basis of V by adding some elements to
  7. a subset S of a vector space V is linearly independent if and only if there exists exactly one linear combination of elements of S which is equal to 0, the one with all zero coefficients.
  8. If a subset S of a vector space V contains 0 then S is linearly dependent.
  9.  A set S with exactly two vectors is linearly dependent if and only if one of these vectors is a scalar multiple of the other.
  10.  Let f1, f2,...,fn be functions in C[0,1] each of which has first n-1 derivatives. The determinant of the following matrix

[ f1(x)

f2(x)

....

fn(x) ]

[ f'2(x)

f'1(x)

....

f'n(x) ]

[ f''2(x)

f''1(x)

....

f''n(x) ]

.....................................

[ f2n-1(x)

f1n-1(x)

....

fnn-1(x) ]


is called the Wronskian of this set of functions.

If the Wronskian of this set of functions is not identically zero then the set of functions is linearly independent (The CONVERSE is not TRUE!!!).

Fact  9(about Linear Transformations):

1. If T is a linear transformation from V to W then T(0)=0.

2. If T is a linear transformation from V to W and S is a linear transformation from W to Y (V, W, Y are vector spaces) then the product (composition) ST is a linear transformation from V to Y.

3. If T and S are linear transformations from V to W (V and W are vector spaces) then the sum T+S which takes every vector A in V to the sum T(A)+S(A) in W is again a linear transformation from V to W.

4. If T is a linear transformation from V to W and k is a scalar then the map kT which takes every vector A in V to k times T(A) is again a linear transformation from V to W.

5. If T is a linear transformation from Rm to Rn then dim(range(T))+dim(kernel(T))=m.

6. If  [T]B,K  is the matrix of a linear  Transformation T  from a vector space V   into  W, then for every vector v in V we have:
                                             [T(v)]K = [T]B,K[v]B.
 

 

Fact 10 (dim(column space of A ) + dim(Nul(A) = number of  columns of A):

  1. For every matrix A the dimension of the column space plus the dimension of the null space is equal to the number of columns of A.

      2.   The column rank and the row rank of every matrix coincide and they  are equal to the (row) rank of the reduced row echelon form of this matrix, and are equal to the number of leading 1's in this reduced row echelon form.

Fact 11(about orthogonal vectors):

1. Every set of pairwise orthogonal non-zero vectors is linearly independent.

 

Fact 12 (about Eigenvalues,  Eigenspaces, and Diagolization )

Let A   be   an  n by  n  matrix. Then
(a)
The characteristic polynomial of  A   is     the     Det (A - xI_n)  and   it   is a polynomial $ p(\lambda)$ of degree $ n$.
(b)
The eigenvalues of $ A$ are the solutions of $ p(\lambda) = 0$.
(c)
If $ \lambda_0$ is an eigenvalue, any nontrivial solution of $ (A-\lambda_0I)X = 0$ is an eigenvector of $ A$ corresponding to $ \lambda_0$.
(d)    $ \lambda_0$  is an eigenvalue  of  A   if and only if   Det  (A - $ \lambda_0$I) = 0. (Where  I  is the identity matrix  n  by n)

(e)  If $ A$ and $ B$ are similar, then the characteristic polynomial of  A  =   the characteristic polynomial of  B, and  hence the set of eigenvalues of $ A$ is equal to the set of eigenvalues of $ B$.

(f)  If $ A$ and $ B$ are similar , i.e.  $ A
= P^{-1}BP$ and $ (\lambda_0,X)$ is an eigenpair of $ A$, then $ (\lambda_0,PX)$ is an eigenpair of $ B$.

(g)  (Diagonalization):   If $ A_{n\times n}$ is diagonalizable, then $ A$ has $ n$ linearly independent eigenvectors. Also, in the equation $ A = PDP^{-1}, P$ is a matrix whose columns are eigenvectors, and the diagonal entries of $ D$ are the eigenvalues corresponding column by column to their respective eigenvectors.

 

 

 

 

 

 


 


 

Diagonalization:


Algorithm:

To diagonalize an n by n matrix A:

  1. First solve the eigenvalue problem for A..
    • If there are fewer than n linearly independent eigenvectors, then A is not diagonable. Stop.
  2. Write down the matrices U, whose columns are the eigenvectors, and D, which has the eigenvalues on the main diagonal and zeroes everywhere else. (The order of the eigenvalues in D must be the same as the order of the eigenvectors in U.)
  3. Write

A = U D U-1.

If a square matrix is not diagonable, it is still possible, and useful, to convert it with a similarity transformation into upper triangular echelon form. In a course on linear algebra you will study this in more depth. Here we will just give an example showing what can happen:

Example 5. Let

     [ 0  1  1 ]
A := [12 -2 -8 ]
     [0   2  2 ]
 

This matrix has a hidden special property. If you calculate its cube, you find, somewhat surprisingly, that

 3   [ 0  0  0 ]
A  = [ 0  0  0 ]
     [ 0  0  0 ]
 

Now, what could the eigenvalues of such a nilpotent matrix be? If Avv, then

A3 v = 3v

and we conclude that the only possible eigenvalue is 0. But if a basis of three eigenvectors all have eigenvalue zero, then the whole matrix A must be the zero matrix. It isn't.


Let's illustrate why diagonalizing is useful with another


Model problem.  Let

     [ 3 -5]
A =  [-5  3].
 
 

Find A10.


Solution. Don't just multiply! Instead, first diagonalize:

A = U-1D U,

where by following the algorithm we find:

1 = -2, with eigenvector

     [ 1]
v =  [ 1].
 1
 

and
1 = 8, with eigenvector

     [ 1]
v =  [-1].
 2
 

Therefore,

     [1/2  1/2] [2  0] [ 1   1]          -1
A =  [1/2 -1/2] [0  8] [ 1  -1] =: U D U  .
 
 

Now, we notice that the product of A with itself p times collapses:

Ap = U D U-1 U D U-1 ... U D U-1

        = U D D ...D U-1,

because U-1 U = I_n. Therefore the answer is:

 6   [1/2  1/2] [26  0] [ 1   1]    [131104  -131040]
A =  [1/2 -1/2] [0  86] [ 1  -1] =  [-131040  131104].
 
 
 
 
 
Now we state and prove the theorem which the last example illustrates.
 

 

Tests for diagonalization: :

An n by n matrix A can be diagonalized if any of the following conditions can be verified:

  1. if all the eigenvalues are different; or 
  2. if A is "symmetric", that is, AT = A; or
  3. if A is "normal", that is, A AT = AT A.

 

 

Matrix Representation of a Linear Transformation :

 

 

List of concepts and definitions

  1. System of linear equations
  2. Solution of a system of linear equations
  3. A consistent system of equations
  4. General solution of a system of linear equations
  5. Augmented matrix of a system of linear equations
  6. Row operations on matrices
  7. Equivalent Matrices
  8. Row-echelon form
  9. Reduced row-echelon form  
  10. Gauss-Jordan elimination procedure
  11. Homogeneous system of linear equations Sum of matrices
  12. Multiplying a matrix by a scalar
  13. Product of matrices
  14. Identity matrix
  15. Transpose of a matrix
  16. Trace of a matrix
  17. Invertible matrix
  18. Inverse of a matrix
  19. Elementary matrix
  20. Row-Equivalent matrices
  21. Symmetric matrices
  22. Diagonal matrices
  23. Upper- and  lower-triangular matrices
  24. Skew-symmetric matrices
  25. Cofactor of entry aij.
  26. Determinant of a square matrix  
  27. Cofactor expansion along a row (column)
  28. The adjoint of a matrix
  29. Cramer's rule
  30. Vector space
  31. Subspace of a vector space.
  32. Linear combination of vectors
  33. Vector space spanned by a set of vectors
  34. Linearly independent sets of elements in a vector space.
  35. Linearly dependent sets of elements in a vector space.
  36. A basis of a vector space
  37. The dimension of a vector space
  38. The null space of a matrix
  39. Nullity of a matrix
  40. Rank of a matrix
  41. Row Space of a Matrix
  42. Column space of a matrix
  43. The Wronskian of a set of functions
  44. Linear transformations
  45. Kernel and Range of a linear transformation
  46. Coordinates of a vector in a given basis
  47. Standard matrix representation of a linear transformation
  48. Eigenvector,  eigenvalue,  eigenspace,  and the characteristic polynomial of a matrix
  49. Similar matrices
  50. Diagonalizable Matrix
  51. Orthogonal vectors
  52. Orthogonal basis
  53. Gram-Schmidt procedure

 

 

 

 

 

 

Definitions in Linear Algebra:   (for examples click here)

A system of linear equations is any sequence of linear equations. A solution of a system of linear equations is any common solution of these equations. A system is called consistent if it has a solution. A general solution of a system of linear equations is a formula which gives all solutions for different values of parameters.
 

Examples. 1. Consider the system:
 

                                          x +  y =  7
                                         2x + 4y = 18

This system has just one solution: x=5, y=2. This is a general solution of the system.
 

2. Consider the system:
 

                                       x +  y + z =  7
                                      2x + 4y + z = 18.

This system has infinitely many solutions given by this formula:
 

                                        x = 5 - 3s/2
                                        y = 2 + s/2
                                        z = s

This is a general solution of our system.

Two matrices are said to be row-equivalent if one can be obtained from the other by a sequence (order may vary) of elementary row operations given below.

(Also see  row-equivalent (2))

* Interchanging two rows 

* Multiplying a row by a non zero constant

* Adding a multiple of a row to another row

 

Row-Echelon Form and Reduced Row-Echelon Form

A matrix in row-echelon form has the following properties.

1.   All rows consisting entirely of zeros occur at the bottom of the matrix
2.   For each row that does not entirely consist of zeroes, the first non zero entry is 1 called a leading 1.
3.   For two successive non zero rows, the leading 1 in the higher row is farther to the left that the leading 1 in the lower row.

A matrix in row-echelon form is in reduced row-echelon form  if every column that has a leading 1 has zeros in every position above and below its leading 1.

 


Augmented Matrix

A matrix derived from a system of linear equations is called the augmented matrix of the system.

Example Consider the system of equations:
 

                       x1 + 2x2 +         x4   =  6
                                     x3 +  6x4  =  7
                                                 x5=1

Its augmented matrix is
 

[ 1 2 0 1 0 6 ]
[ 0 0 1 6 0 7 ]
[ 0 0 0 0 1 1 ]

The matrix has the reduced row echelon form. The leading unknowns are x1, x3 and x5; the free unknowns are x2 and x4. So the general solution is:

                              x1= 6-2t-s
                              x2= s
                              x3= 7-6t
                              x4= t
                              x5= 1

If the augmented matrix does not have the reduced row echelon form but has the (ordinary) row echelon form then the general solution also can be easily found.
 

The method of finding the solution is called the back-substitution.
 

First we solve each of the equations for the leading unknowns The last non-zero equation gives us the expression for the last leading unknown in terms of the free unknowns. Then we substitute this leading unknown in all other equations by this expression. After that we are able to find an expression for the next to the last leading unknown, replace this unknown everywhere by this expression, etc. until we get expressions for all leading unknowns. The expressions for leading unknowns that we find in this process form the general solution of our system of equations.
 

Example. Consider the following system of equations.
 

                    x1-3x2+ x3-x4 = 2
                         x2+2x3-x4 = 3
                              x3+x4 = 1

Its augmented matrix
 

[ 1 -3 1 -1 2 ]
[ 0 1 2 -1 3 ]
[ 0 0 1 1 1 ]

is in the row echelon form.
 

The leading unknowns are x1, x2, x3; the free unknown is x4.

Solving each equation for the leading unknown we get:

                            x1=2+3x2-x3+x4
                            x2=3-2x3+x4
                            x3=1-x4

The last equation gives us an expression for x3: x3=1-x4. Substituting this into the first and the second equations gives:
 

                    x1=2+3x2-1+x4+x4=1+3x2+2x4
                    x2=3-2(1-x4)+x4=1+3x4
                    x3=1-x4

Now substituting x2=1+3x4 into the first equation, we get
 

                    x1=1+3(1+3x4)+2x4=4+11 x4
                    x2=1+3x4
                    x3=1-x4

Now we can write the general solution:
 

                                    x1=4+11 s
                                    x2=1+ 3 s
                                    x3=1-   s
                                    x4=     s

Let us check if we made any arithmetic mistakes. Take x4=1 and compute x1=15, x2=4, x3=0, x4=1. Substitute it into the original system of equations:
 

                            15 - 3 * 4 + 0 - 1 = 2
                                 4 + 2 * 0 - 1 = 3
                                         0 + 1 = 1

OK, it seems that our solution is correct.

 

 

Consistent System

A system of linear equations which has a solution.

 

Homogeneous System

A homogeneous system of linear equations is of the form:

a11x1 + a12x2 + . . . + a1nxn=0
a21x1 + a22x2 + . . . + a2nxn=0

am1x1 + am2x2 + . . . + amnxn=0

That is, a homogeneous system of linear equations has all constant terms equal to zero.

The Gauss-Jordan elimination procedure

There exists a standard procedure to obtain a reduced row echelon matrix from a given matrix by using the row operations.
This procedure consists of the following steps.
 

  1. Locate the leftmost column which does not consist of zeroes.
  2. If necessary swap the first row with the row which contains a non-zero number a in the column found on step 1.
  3. If this number a is not 0, multiply the first row by 1/a, to get a leading 1 in the first row.
  4. Use the first row to make zeroes below the leading 1 in the first row (by using the adding operation).
  5. Cover the first row and apply the first 4 steps to the remaining sub-matrix. Continue until the whole matrix is in the row echelon form.
  6. Use the last non-zero row to make zeroes above the leading 1 in this row. Use the second to last non-zero row to make zeroes above the leading 1 in this row. Continue until the matrix is in the reduced row echelon form.

 

Row Equivalent Matrices

Two matrices A and B are row equivalent if A is a product of elementary matrices times B, or equivalently if A can be obtained from B by a finite sequence of elementary row operations.

In order to multiply a matrix by a scalar, one has to multiply all entries of the matrix by this scalar.
 

Example:

3 *
[ 2 3 1 2 ]
[ 3 1 0 1 ]
[ 1 2 2 3 ]
=
[ 6 9 3 6 ]
[ 9 3 0 3 ]
[ 3 6 6 9 ]

 

Pruduct of two matrices:

The product of a row-vector v of size (1, n) and a column vector u of size (n,1) is the sum of products of corresponding entries:

uv=u(1)v(1)+u(2)v(2)+...+u(n)v(n)
 

Example:

                        3
            (1, 2, 3) * 4 =1*3 + 2*4 + 3*1 = 3+8+3=14
                        1

Example:

                        x
            (2, 4, 3) * y = 2x + 4y + 3z
                        z

As you see, we can represent the left side of a linear equation as a product of two matrices. The product of arbitrary two matrices which we shall define next will allow us to represent the left side of any system of equations as a product of two matrices.
 

Let A be a matrix of size (m,n), let B be a matrix of size (n,k) (that is the number of columns in A is equal to the number of rows in B. We can subdivide A into a column of m row-vectors of size (1,n). We can also subdivide B into a row of k column-vectors of size (n,1):
 

          r1
          r2
       A =...      B=[c1 c2 ... ck]
          rm

Then the product of A and B is the matrix C of size (m,k) such that
 

C(i,j)=ricj
 

(C(i,j) is the product of the row-vector ri and the column-vector cj).
 

 

 

Matrices A and B such that the number of columns of A is not equal to the number of rows of B cannot be multiplied.
 

Example:
 

[ 2 3 4 ]
[ 1 2 3 ]
*
[ 1 2 ]
[ 3 4 ]
5 6
=
[ 31 40 ]
[ 22 28 ]

Example:
 

[ 2 3 4 ]
[ 1 2 3 ]
*
[ x ]
[ y ]
[ z ]
=
[ 2x + 3y + 4z ]
[ x + 2y + 3z ]


 

You see: we can represent the left part of a system of linear equations as a product of a matrix and a column-vector. The whole system of linear equations can thus be written in the following form:

AX = b

Identity Matrix I_n:

Let In denote the identity matrix of order n that is a square matrix of order n with 1s on the main diagonal and zeroes everywhere else. Then for every m by n matrix A the product of A and In is A and the product of Im and A is A.

 

Nonsingular (invertible) matrices:

A square matrix of size n A is called invertible (nonsingular)  if there exists a square matrix B of the same size such that AB = BA = In, the identity matrix of size n. In this case B is called the inverse of A and B is denoted by A-1.

 

FINDING THE INVERSE OF A MATRIX

Let A be square matrix order n.

1.   Write a matrix that consists of the given matrix A on the left and the identity matrix n x n next to A on the right, separating the matrices A and I by a (vertical) dotted line to obtain [A:I].

2.  If possible, row reduce A to I using elementary row operations on the matrix [A:I]. The result will be the matrix [I:A-1].
Note that for some matrices an inverse does not exist. If it is not possible to obtain [I:A-1], then the matrix A is not invertible.

3.   Check your work by multiplying to see that AA-1 = I = A-1 A.

 

Elementary matrices.

An elementary matrix is a matrix obtained from an identity matrix by one of the row-operations.

Example.

[ 1

3 ]

[ 0

1 ]

,

[ 1

0

0 ]

[ 1

1

0 ]

[ 0

0

1 ]

,

[ 1

0

0 ]

[ 0

0

1 ]

[ 0

1

0 ]

,

[ 1

0

0 ]

[ 0

2

0 ]

[ 0

0

1 ]

The first two matrices are obtained by adding a multiple of one row to another row. The third matrix is obtained by swapping two rows. The last matrix is obtained by multiplying  row by a number.

As we see, elementary matrices usually have lots of zeroes.

Elementary matrices which are obtained by adding rows contain only one non-diagonal non-zero entry.

Elementary matrices which are obtained by multiplying a row by a number contain exactly 1 non-unit entry on the diagonal and no non-zero entries outside the diagonal.

Elementary matrices which are obtained by swapping consist of 0s and 1s and contain exactly 2 non-diagonal entries.

The converse statements are true also (for example every matrix with 1s on the diagonal and exactly one non-zero entry outside the diagonal) is an elementary matrix.

The main result about elementary matrices is that every invertible matrix is a product of elementary matrices. These are in some sense the smallest particles in the world of invertible matrices.

 

Transpose of a matrix:

If A is any m by n matrix then the transpose of A, denoted by AT, is defined to be the n by m matrix obtained by interchanging the rows and columns of A, that is the first column of AT is the first row of A, the second column of AT is the second row of A, etc.
 

Example.
The transpose of

[ 2 3 ]
[ 4 5 ]
[ 6 7 ]
is
[ 2 4 6 ]
[ 3 5 7 ]

Trace of a Matrix:

If A is a square matrix of size n then the sum of the entries on the main diagonal of A is called the trace of A and is denoted by tr(A).

Example.
The trace of the matrix

[ 1 2 3 ]
[ 4 5 6 ]
[ 7 8 9 ]

is equal to 1+5+9=15

 

Symmetric, diagonal, triangular matrices

A square matrix A is called symmetric if AT=A, that is if A(i,j)=A(j,i) for every i and j. Thus A is symmetric if and only if it is symmetric with respect to the main diagonal. Here is an example of a symmetric matrix:

[ 1

2

3 ]

[ 2

4

5 ]

[ 3

5

6 ]

An important subclass of symmetric matrices is formed by diagonal matrices, i.e. matrices which have zeroes everywhere outside the main diagonal. For example, the identity matrix is a diagonal matrix.

 

A square matrix A is called upper triangular (resp. lower triangular) if all its entries below (resp. above) the main diagonal are zeroes, that is A(i,j)=0 whenever i is greater than j (resp. A(i,j)=0 whenever i is less than j).

Example.

[ 1

2

3 ]

[ 0

0

1 ]

[ 0

0

1 ]

,

[ 1

0

0 ]

[ 0

1

0 ]

[ 2

3

4 ]

The first of these matrices is upper triangular, the second is lower triangular.

A square matrix A is called skew-symmetric if A^T = -A, that is A(i,j)=-A(j,i) for every i and j. In particular, if i=j then A(i,i)=0, that is the diagonal entries of a skew-symmetric matrix are equal to 0.

Example:

[ 0

2

3 ]

[ -2

0

4 ]

[ -3

-4

0 ]

 

Determinant:

Let Aij is the cofactor of entry  a_ij  that is  equal to  (-1)i+j multiplied by the determinant of the matrix obtained from A by deleting the i-th row and the j-th column (the determinant of this smaller matrix has a special name also; it is called the minor of entry A(i,j)).

  1. For every i=1,...,n we have det(A)=a_i1 Ai1 + a_i2 Ai2 +...+ a_in Ain This is called the cofactor expansion of det(A) along the i-th row (a similar statement holds for columns).

  2. The determinant of a triangular matrix is the product of the diagonal entries.

Adjoint Matrix:

Definition. If A is any n by n matrix and Aij is the cofactor of a_ij then the matrix

[ A11

A21

...

An1 ]

[ A12

A22

...

An2 ]

............................

[ A1n

A2n

...

Ann ]

is called the adjoint of A, denoted adj(A). (Notice that this matrix is obtained by replacing each entry of A by its cofactor and then taking the transpose.)

Cramer's rule

(Cramer's rule) If A is an invertible matrix then the solution of the system of linear equalitions
AX=b
 

can be written as
x1=det(A1)/det(A),  x_2 = det(A_2)/Det(A)...
...,
xn= det(An)/det(A)


 

where Ai is the matrix obtained from A by replacing the i-th column by the column vector b.
 

VECTOR SPACE :

Any set of objects V where addition and scalar multiplication are defined and satisfy properties 1--7 is called a vector space.  Let  A, B, C  be elements  in V, and a, b  be real numbers:

  1. The addition is commutative and associative: A+B=B+A, A+(B+C)=(A+B)+C

  2. The multiplication by scalar is distributive with respect to the addition: a(B+C)=aB+aC

  3. The product by a scalar is distributive with respect to the addition of scalars: (a+b)C=aC+bC

  4. a(bC)=(ab)C

  5. 1*A=A (here 1 is the scalar 1)

  6. There exists a zero n-vector 0 such that 0+A=A+0=A for every A

  7. 0*A=0 (here the first 0 is the scalar 0, the second 0 is the zero-vector)

Continuous functions, matrices, R^n, P_n,  and (complex numbers )satisfy the abave properties.
 

Of course, we denote B+(-A) as B-A, so the subtraction is a derived operation, we derive it from the addition and the multiplication by scalar. Notice that in the solution of the equation X+A=B we did not use the fact that A,B,X are functions or vectors (or matrices or numbers for that matter), we used only the properties of our basic operations.
 

Here by addition we mean any operation which associates with each pair of objects A and B from V another object (the sum) C also from V; by a scalar multiplication we mean any operation which associates with every scalar k and every object A from V another object from V called the scalar multiple of A and denoted by kA.

Elements of general vector spaces are usually called vectors. For any system of vectors A1,...,An and for any system of numbers a1,...,an one can define a linear combination of A1,...,An with coefficients a1,...,an as
a1A1+...+anAn.

Notice that a vector space does not necessarily consist of n-vectors or vectors on the plane. The set of continuous functions, the set of k by n matrices, the set of complex numbers are examples of vector spaces.

Not every set of objects with addition and scalar multiplication is a vector space. For example, we can define the following operations on the set of 2-vectors:
 

Addition: (a,b)+(c,d)=(a+c,d).
 

Scalar multiplication: k(a,b)=(k2a,b).
 

Then the resulting algebraic system will not be a vector space because if we take k=3, m=2, a=1, b=1 we have:
 

(k+m)(a,b)=(3+2)(1,1)=5(1,1)=(25,1); k(a,b)+m(c,d)=3(1,1)+2(1,1)=(9,1)+(4,1)=(13,1),

Thus the third property of vector spaces does not hold.

Subspaces

Let V be a  vector space . A subset W of V is called a subspace of V if W is closed under addition and scalar multiplication, that is if for every vectors A and B in W the sum A+B belongs to W and for every vector A in W and every scalar k, the product kA belongs to W.

Positive Examples.

1. The whole space Rn is a subspace of itself. And the set consisting of one vector, 0, is a subspace of any space.

2. In R2, consider the set W of all vectors which are parallel to a given line L. It is clear that the sum of two vectors which are parallel to L is itself parallel to L, and a scalar multiple of a vector which is parallel to L is itself parallel to L. Thus W is a subspace.

3. A similar argument shows that in R3, the set W of all vectors which are parallel to a given plane (line) is a subspace.

4. The set of all polynomials is a subspace of the space of continuous functions on [0,1], C[0,1]. The set of all polynomials whose degrees do not exceed a given number, is a subspace of the vector space of polynomials, and a subspace of C[0,1].

5. The set of differentiable functions is also a subspace of C[0,1].

Negative Examples.

1. In R2, the set of all vectors which are parallel to one of two fixed non-parallel lines, is not a subspace. Indeed, if we take a non-zero vector parallel to one of the lines and add a non-zero vector parallel to another line, we get a vector which is parallel to neither of these lines.

2. The set of polynomials of degree 2 is not a subspace of C[0,1]. Indeed, the sum of x2+x and -x2 is a polynomial of degree 1.


Linear Combination

A linear combination of the vectors v1, v2, . . . ,vn  is a vector w of the form

w = α1v1 + α2v2 + . . . + αnvn .

 

Spanning set:

Let v1, v2, . . . ,vn  be elements of a vector space  V.  Then  Span{ v1, v2, . . . ,vn }  is the set ofall possible  linear combination of the elements  v1, v2, . . . ,vn . It is easily verified that   Span{ v1, v2, . . . ,vn }  is a subspace of  V  (i.e.  Span{ v1, v2, . . . ,vn } is a vector space).

Linearly Independent , Linearly Dependent:

A subset S of a vector space V is called linearly independent if no element of S is a linear combination of other elements of S.
 


A subset S of a vector space is called linearly dependent if it is not linearly independent.

Notice that if a subset of a set S is linearly dependent then the whole set is linearly dependent (the proof is an exercise).

Examples.
1. Every two vectors in the line R are linearly dependent (one vector is proportional to the other one).
 

2. Every three vectors on a plane R2 are linearly dependent. Indeed, if two of these vectors are parallel then

3. If a subset S of Rn consists of more than n vectors then S is linearly dependent. ( Prove it!)
 

4. The set of polynomials x+1, x2+x+1, x2-2x-2, x2-3x+1 is linearly dependent. To prove that we need to find numbers a, b, c, d not all equal to 0 such that
a(x+1)+b(x2+x+1)+c(x2-2x-2)+d(x2-3x+1)=0
This leads to a homogeneous system of linear equations with 4 unknowns and 3 equations. Such a system must have a non-trivial solution by a result  about homogeneous systems of linear equations.

Wronskian

Let f1, f2,...,fn be functions in C[0,1] each of which has first n-1 derivatives. The determinant of the following matrix
 

[ f1(x) f2(x) .... fn(x) ]
[ f'2(x) f'1(x) .... f'n(x) ]
[ f''2(x) f''1(x) .... f''n(x) ]
.....................................
[ f2n-1(x) f1n-1(x) .... fnn-1(x) ]


is called the Wronskian of this set of functions.
 

Theorem. Let f1, f2,...,fn be functions in C[0,1] each of which has first n-1 derivatives. If the Wronskian of this set of functions is not identically zero then the set of functions is linearly independent.
 

 


 

Basis, Rank,  and dimension:

A set S of vectors in a vector space V is called a basis if

  1. S is linearly independent and
  2. V=span(S), that is every vector in V is a linear combination of vectors in S.

A dimension of a vector space V (denoted by dim(V) ), is the number of elements in a basis for V. There is one exception of this definition: the dimension of the zero space (the vector space consisting of one vector, zero) is defined to be 0 and not 1.

Let S be a set of vectors in a vector space. Then the dimension of the vector space spanned by S is called the rank of S.

If A is a matrix then the rank of the set of its columns is called the column rank of A and the rank of the set of its rows is called the row rank of A.

It is known that the column rank of A is equal to the row rank of A for every matrix a.

Column Space

The column space of an mxn matrix A is the span of the columns of A and is denoted by col(A).

Row Space

The row space of a matrix A is the span of the rows of A and is denoted by row(A).

Nullspace (or Kernel)

The nullspace of an mxn matrix A is null(A) = {x Rn     such that    Ax = 0}.

Nullity

The nullity of a matrix A is the dimension of the nullspace of A.

nullity(A) = dim(null(A))

Linear transformations of arbitrary vector spaces

Let V and W be arbitrary vector spaces. A map T from V to W is called a linear transformation if

  1. For every two vectors A and B in V

T(A+B)=T(A)+T(B);

 

  1. For every vector A in V and every number k

T(kA)=kT(A).

 

In the particular case when V=W, T is called a linear operator in V.

Positive examples.1. Let V be the set of all polynomials in one variable. We shall see later that V is a vector space with the natural addition and scalar multiplication (it is not difficult to show it directly). The map which takes each polynomial to its derivative is a linear operator in V as easily follows from the properties of derivative:

(p(x)+q(x))' = p'(x) +q'(x),
(kp(x))'=kp'(x)
.

 

2. Let C[0,1] be the vector space of all continuous functions on the interval [0,1]. Then the map which takes every function S(x) from C[0,1] to the function h(x) which is equal to the integral from 0 to x of S(t) is a linear operator in C[0,1] as follows from the properties of integrals.

int(T(t)+S(t)) dt = int T(t)dt + int S(t)dt
int kS(t) dt = k int S(t) dt.

 

3. The map from C[0,1] to R which takes every function S(x) to the number S(1/3) is a linear transformation (1/3 can be replaced by any number between 0 and 1):

(T+S)(1/3)=T(1/3)+S(1/3),
(kg)(1/3)=k(S(1/3)).

 

4. The map from the vector space of all complex numbers C to itself which takes every complex number a+bi to its imaginary part bi is a linear operator (check!).

5. The map from the vector space of all n by n matrices (n is fixed) to R which takes every matrix A to its (1,1)-entry A(1,1) is a linear transformation (check!).

6. The map from the vector space of all n by n matrices to R which takes every matrix A to its trace trace(A) is a linear transformation .

7. The map from an arbitrary vector space V to an arbitrary vector space W which takes every vector v from V to 0 is a linear transformation . This transformation is called the null transformation

8. The map from an arbitrary vector space V to V which takes every vector to itself (the identity map) is a linear operator (check!). It is called the identity operator, denoted I.

Negative examples. 1. The map T from which takes every function S(x) from C[0,1] to the function S(x)+1 is not a linear transformation because if we take k=0, S(x)=x then the image of kT(x) (=0) is the constant function 1 and k times the image of T(x) is the constant function 0. So the second property of linear transformations does not hold.

 

kernels and ranges of linear transformations

Sources of subspaces: kernels and ranges of linear transformations

Let T be a linear transformation from a vector space V to a vector space W. Then the kernel of T is the set of all vectors A in V such that T(A)=0, that is

ker(T)={A in V | T(A)=0}

The range of T is the set of all vectors in W which are images of some vectors in V, that is

range(T)={A in W | there exists B in V such that T(B)=A}.

Notice that the kernel of a transformation from V to W is a subspace of V and the range is a subspace of W. For example if T is a transformation from the space of functions to the space of real numbers then the kernel must consist of functions and the range must consist of numbers.

 

Coordinate of a vector in a given basis:

Let S be a basis of a vector space V and let a be a vector from V. Then a can be  uniquely represented  as a linear combination of elements from S. The coefficients of  this linear combination are called the coordinates of a in the basis S, we use  the notation  [a]S To denote the coordinates of a in the basis S

Example.
Vectors (1,2) and (2,3) form a basis of R2 . The vector (4,7) is equal to the linear combination 2(1,2)+(2,3). Thus the vector (4,7) has coordinates 2, 1 in the basis of vectors (1,2) and (2,3),  we  write [(4, 7)]S  = [2  1]. The same vector has coordinates 4 and 7 in the basis of vectors D = (1,0) and (0,1). Thus a vector has different coordinates in different bases. It is sometimes very important to find a basis where the vectors you are dealing with have the simplest possible coordinates.

 

 

Matrix representation of a linear Transformation

Let  B={b1,...,bn} be a basis in an n-dimensional vector space V and  K = {d1,...,dm} be a basis in an m-dimensional vector space W . For every vector v in V let [v]B denote the column vector of coordinates of v in the basis B.
 

Let T be a linear Transformation from  V  into  W.  Since T(b1),...,T(bn) are in W, each of these vectors is a linear combination of vectors from K. Consider the m by n matrix [T] B,K  whose columns are the column vectors of coordinates [T(b1)]K,..., [T(bn)]K. This matrix is called the matrix of the linear operator T in the basis B and K.

 

Important Result :  If [T]B,K is the matrix of a linear  Transformation T from a vector space V   into  W, then for every vector v in V we have:
 

[T(v)]K = [T]B,K*[v]B.
 

 

 

Characteristic Polynomial

The characteristic polynomial of a square matrix A is defined by
p(x) = det(xI - A).

 

 

Eigenvalue  The scalar λ is said to be an eigenvalue of an nxn matrix A if there is a nonzero vector x with Ax = λx , the eigenvalues of a matrix   A, n by n,   are found  by setting  the characteristic polynomial of  A  equals to  zero.

A non-zero vector w is called an eigenvector of a matrix A with the eigenvalue a if Aw=aw (and hence a is an eigenvalue of A that corresponds to the eigenvector w.)
 

We can rewrite the equality Aw=aw in the following form:
 

Aw=aIw
 

where I is the identity matrix. Moving the right hand side to the left, we get:
 

(A-aI)w=0
 

This is a homogeneous system of linear equations. If the matrix A has an eigenvector W then this system must have a non-trivial solution (w is not 0 !). this is equivalent to the condition that the matrix A-aI is not invertible, and  this is equivalent to the condition that det(A-aI) is  equal to zero. Thus a matrix A has an eigenvector with an eigenvalue a if and only if det(A-aI)=0.


 

Notice that det(A-xI) is a polynomial with the unknown x. This polynomial is called the characteristic polynomial of the matrix A.
 

Thus in order to find all eigenvectors of a matrix A, one needs to find the characteristic polynomial of A, find all its roots (=eigenvalues) and for each of these roots a, find the solutions of the homogeneous system (A-aI)w=0.
 

Example.
Let A be the matrix
 

[ 1 2 ]
[ 2 1 ]

The characteristic polynomial det(A-xI) is (1-x)2-4. The roots are x=-1 and x=3. These are the eigenvalues A.
 

Let a=-1. Then the system of equations (A-aI)w=0 has the form
 

2x+2y=0
2x+2y=0
 

where w=(x,y). The general solution is:
 

x=-t
y=t

 

This solution gives the set of all eigenvalues with the eigenvalue -1. It is a subspace spanned by the vector b1=(-1,1).
 

Now let a=3. Then the system (A-aI)w=0 has the form
 

-2x+2y=0
2x-2y=0
 

The general solution is:
 

                                                x = t 

                                                y = t

Thus the set of eigenvectors with eigenvalue 3 forms a subspace spanned by the vector (1,1).

Eigenvector : The nonzero vector x is said to be an eigenvector of the matrix A, associated with the eigenvalue λ if Ax = λx.    Thus in order to find all eigenvectors of a matrix A, one needs to find the characteristic polynomial of A, find all its roots (=eigenvalues) and for each of these roots a, find the solutions of the homogeneous system (A-aI)v=0.

Eigenspace : The eigenspace associated with the eigenvalue λ of a matrix A is defined by
Eλ = null(λI - A).

 

Similar Matrices

Matrices A and B are said to be similar if there exists an invertible matrix S such that
A = S-1BS.

Diagonalizable Matrix

A square matrix A is said to be diagonalizable if there is an invertible matrix P so that

P-1AP is a diagonal matrix.

Orthogonal Vectors

The vectors u and v in Rn are said to be orthogonal if uv (u dot product  v)  = 0.

The vectors u and v in an inner product space are said to be orthogonal if <u,v> = 0.

Notation for both of these is u v.

Orthonormal Vectors

Vectors are said to be orthonormal if they are pairwise orthogonal and are all unit vectors.

Orthogonal Basis

A set of vectors B is said to be an orthogonal basis for a subspace V if:

1) the vectors in B form a basis for V
2) the vectors in the set are pairwise orthogonal

Gram-Schmidt procedure: (Orthogonal bases. The Gram-Schmidt algorithm)

A basis of a  vector space is called orthogonal if the vectors in this basis are pairwise orthogonal.
 

A basis of a Euclidean vector space ( think of  R^n,  and  if  x, y  in  R^n, then  <x, y>  means  the dot product of  x with y, i.e   <x, y> = x..y ) is called orthonormal if it is orthogonal and each vector has norm 1.
 

Given an orthogonal basis {v1,...,vn}, one can get an orthonormal basis by dividing each vi by its length: {v1/||v1||,...,vn/||vn||}.
 

Suppose that we have a (not necessarily orthogonal) basis {s1,...,sn} of a Euclidean vector space V. The next procedure, called the Gram-Schmidt algorithm, produces an orthogonal basis {v1,...,vn} of V. Let
 

v1=s1
 

We shall find v2 as a linear combination of v1 and s2: v2=s2+xv1. Since v2 must be orthogonal to v1, we have:
 

0=v2v1=(s2+xv1)v1=s2v1 + x<v1,v1>.
Hence
 
x=-<s2,v1>/<v1,v1>,
 

so
                                                                                                          v2=s2-(<s2,v1>/<v1,v1>) v1
 

Next we can find v3 as a linear combination of s3, v1 and v2. A similar calculation gives that
 

v3=s3 - (<s3,v1>/<v1,v1>) v1 - (<s3,v2>/<v2,v2>) v2.
 

Continuing in this manner, we can get the formula for vi+1:
 

vi=si - (<si,v1>/<v1,v1>) v1 - (<si,v2>/<v2,v2>) v2-...-(<si,vi-1>)/<vi-1,vi-1>) vi-1
 

By construction, the set of vectors {v1,...,vn} is orthogonal. None of these vectors is 0. Indeed, if vi were equal to 0 then si, v1,...,vi-1 would be linearly dependent, which would imply that s1,...,si are linearly dependent (replace each vj as a linear combination of s1,...,sj), which is impossible since {s1,...,sn} is a basis. This implies that {v1,...,vn} are linearly independent. Since {v1,...,vn} is an orthogonal basis , one can get an orthonormal basis by dividing each vi by its length: {v1/||v1||,...,vn/||vn||} is an orthonormal basis.