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Eigenvalues and eigenvectors
Denition 10.1 The scalar R is called an eigenvalue of the matrix A Rn×n
if there exists a nonzero vector s Rn for which
As = s

holds. The vector s is said to be the eigenvector (corresponding to the eigenvalue ) of matrix A. matrix

Remark 10.1 If both s and t are eigenvectors corresponding to the eigenvalue
then their any nontrivial linear combination s + t is also an eigenvector with the same eigenvalue :
A (s + t) = Ax + At = s + t = (s + t) .

The eigenvectors corresponding to the eigenvalue are the nonzero solutions of the homogeneous system of linear equations (A - E) x = 0. A homogeneous system of linear equations with a square coecient matrix as it is already known has nonzero solution if and only if its coecient matrix is singular. Furthermore a square matrix is singular if and only if its determinant is zero: |A - tE| = 0 It follows that is an eigenvalue of A if and only if it satises the above, so called characteristic equation.

Denition 10.2 The characteristic polynomial of a matrix A Rn×n
p(t) = |A - tE|

where E is the identity matrix of order n.
Therefore we can also state: is an eigenvalue of A if and only if it is a root of its characteristic polynomial.

Example 10.1 Find the eigenvalues and the corresponding eigenvectors of the following matrix:
1 A= 0 2
Solution:



0 1 0

-1 1 -2

Now there are two ways to nd the eigenvalues and the corresponding eigenvectors: 1) We nd the values of the parameter t for which the matrix A - tE is singular and with the obtained values we nd the nonzero solutions of the homogeneous system of linear equations (A - tE)x = 0, or 1

2) We determine the roots of the characteristic polynomial of A, those are the eigenvalues of A and then solving the homogeneous system of linear equations (A - E) x = 0 for each root we obtain the corresponding eigenvectors. 1st method: Using elementary basis transformation we solve the homogeneous system of linear equations (A - tE)x = 0 for each value of the parameter t for which the coecient matrix is singular:

x1 1-t 0 2

x2 x3 x 0 -1 3 1-t 1 0 -2 - t x1 0 0 2 x2 0 x 3 0 x1 0

x1 x2 t-1 0 1-t 1-t t2 + t 0

If t = 1 then we can continue as follows:

x3

x2 0 0 0

From this table we obtain that at t = 1 the coecient matrix is singular and the solution of the homogeneous system of linear equations (A - E)x = 0 x1 0 x2 = 1 R , x3 0 that is 1 is an eigenvalue and the above vectors are the corresponding eigenvectors for all = 0. If t = 1 then x1 x2 x1 x3 t - 1 0 x t-1 3 x2 1 1-t 1-t t2 + t t2 + t 0 and we obtain that the coecient matrix is also singular if t = 0 or t = -1. It can be red of from the table, that the eigenvectors corresponding to the eigenvalue 0 are x1 1 x2 = -1 R = 0 1 x3 and the the eigenvectors corresponding to the eigenvalue -1 are x1 1 x2 = -1 R = 0. x3 2

2nd method: The characteristic equation:
|A - tE| = 1-t 0 2 0 -1 1-t 1 0 -2 - t = -t3 + t = 0

Its solutions: t1 = 0, t2 = -1 and t3 = 1 are the eigenvalues of the matrix A.

· The eigenvectors belonging to the eigenvalue 0 are the nonzero solutions of the homogeneous system of linear equations (A - 0 · E) x = 0. Thus the system to be solved: 1 0 -1 x1 0 0 1 1 x2 = 0 . 2 0 -2 x3 0
2

The solutions: x1 = , x2 = -, x3 = ( R) . Therefore the eigenvectors belonging to the eigenvalue t1 = 0 are 1 -1 , ( R \ {0}) 1

· The eigenvectors belonging to the eigenvalue -1 are the nonzero solutions of the homogeneous system of linear equations (A + 1 · E) x = 0. Thus the system to be solved: 2 0 -1 x1 0 0 2 1 x2 = 0 . 2 0 -1 x3 0
The solutions: x1 = , x2 = -, x3 = 2 ( R) . Therefore the eigenvectors belonging to the eigenvalue t2 = -1: 1 -1 , ( R \ {0}) . 2

· The eigenvectors belonging to the eigenvalue 1 are the nonzero solutions of the homogeneous system of linear equations (A - 1 · E) x = 0, that is, 0 0 -1 x1 0 0 0 1 x2 = 0 . 2 0 -3 x3 0
Solving the system the eigenvectors : 0 1 , 0

( R \ {0}) .

Remark 10.2 There are real matrices which has no real eigenvalue, for instance the
matrix
A= 0 -1 1 0

has the characteristic polynomial t2 + 1, therefore it has no real root.

Diagonal matrices
A diagonal matrix is a square matrix, whose entries outside of the main diagonal are zeros. They will be denoted by

D = diag d1 , . . . , dn , d1 , . . . , dn are the entries in the main diagonal. The eigenvalues of a diagonal matrix are just the entries in the main diagonal. A square matrix A of order n is said to be diagonalizable if there exists a regular matrix P and a diagonal matrix D such that P-1 AP = D = diag d1 , . . . , dn .

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If P is an arbitrary invertible matrix, then the characteristic polynomial of A and P-1 AP are equal to each other:

P-1 AP - tE

= = =

P-1 AP - tP-1 EP P-1 (A - tE) P P-1 · |A - tE| · |P| = |A - tI|

It follows that; if A is diagonalizable then P-1 AP = diag 1 , . . . , n , where 1 , . . . , n are the eigenvalues of A.

Theorem 10.1 A square matrix A of order n is diagonalizable if and only if it has
n linearly independent eigenvectors x1 , . . . , xn . In this 1 0 . . . 0 0 2 . . . 0 P-1 AP = . . . . . . . . . 0 0 ...

case
,

n

where the columns of P are x1 , . . . , xn , and 1 , . . . , n scalars are the corresponding eigenvalues.
Proof: If x1 , . . . , xn are linearly independent eigenvectors of A and 1 , . . . , n are the corresponding eigenvalues, that is,

Ax1

= 1 x1 . . . = n xn .

Axn

then let P = [x1 , . . . , xn ] . Then it is obtained that

AP

= A [x1 , . . . , xn ] = [Ax1 , . . . , Axn ] = = [1 x1 , . . . , n xn ] [x1 , . . . , xn ] diag 1 , . . . , n

= Pdiag 1 , . . . , n
therefore

P-1 AP = diag 1 , . . . , n .

In the last step we used the fact that P is invertible if and only if its columns x1 , . . . , xn are linearly independent. Conversely, if P-1 AP = diag 1 , . . . , n , then

AP = Pdiag 1 , . . . , n ,

that is, for each column xi (i = 1, . . . , n) of P the equation Axi = i xi holds, verifying that i is an eigenvalue of A and a corresponding eigenvector is xi .

Example 10.2 Decide if the matrix
A= 3 2 1 4

is diagonalizable! If it is diagonalizable then give an invertible matrix P for which P-1 AP is a diagonal matrix.
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Solution: The eigenvalues of A are 1 = 2 and 2 = 5, furthermore the corresponding linearly independent eigenvectors:

v1 =

-1 1

and

v2 =

1 2

Hence by the theorem (10.1) A is a diagonalizable matrix. -1 1 If P = , then 1 2

P-1 AP

= - =

1 3 2 0

2 -1 0 5

-1 -1

3 2

1 4

-1 1

1 2

The necessary and sucient condition for a matrix is being diagonalizable in theorem (10.1) can not be easily checked. Therefore it is an important result can be proved using mathematical induction that eigenvectors of a square matrix A corresponding to dierent eigenvalues are linearly independent, from which it immediately follows that an n × n real matrix A having n dierent real eigenvalues is diagonalizable. Of course the above condition is not necessary one.

Example 10.3 Consider the matrix:
2 A= 0 0 1 1 1 0 0 2

This matrix has only two dierent eigenvalues 1 = 2 and 2 = 1 (where 1 = 2 is a root of the characteristic polynomial with multiplicity two). The linearlyindependent eigenvectors corresponding to the eigenvalue 1 = 2 0 1 are: v1 = 0 and v2 = 0 , while the eigenvector belonging to the eigen1 0 1 value 2 = 1 is v3 = -1 . 1 Thus the matrix has three linearly independent eigenvectors, therefore it is diagonalizable. Hence, if 1 0 1 P = 0 0 -1 , 0 1 1 then P-1 AP =diag 2, 2, 1 . two conditions hold:

Theorem 10.2 A matrix A Rn×n is diagonalizable if and only if the following
1. its system of eigenvalues has n elements if each eigenvalue is taken according to its algebraic multiplicity, 2. the algebraic multiplicity of each eigenvalue equals to the degree of freedom of the homogeneous system of linear equation (A - E) x = 0.

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The square matrices occurring in economical applications are often symmetric, therefore it seems to be reasonable to study the question if the symmetric matrices are diagonalizable. Recall that a real matrix A is said to be symmetric if A = A and it is orthogonal if A = A-1 .

Theorem 10.3
-

- The characteristic equation of an n × n symmetric matrix has n real roots counting their multiplicity as well.

The eigenvectors corresponding to dierent eigenvalues of a symmetric matrix are orthogonal.

Theorem 10.4 The symmetric matrices are diagonalizable, furthermore to each
symmetric matrix A there exists an orthogonal matrix U, such that
U-1 AU = diag 1 , . . . , n ,

where 1 , . . . , n are the eigenvalues of A.

Example 10.4 Find the orthogonal matrix the matrix U AU is a diagonal one. 3 A= 0 0
Solution:

U to the symmetric matrix A for which 0 0 1 0 1 0

The eigenvalues of the matrix are 1 = -1, 2 = 1 and 3 = 3, and the corresponding eigenvectors are 0 1 0 v1 = -1 , v2 = 1 and v3 = 0 . 1 0 1 The normed eigenvectors are the columns of the orthogonal matrix U. Thus



0
1 2

0
1 2 1 2

1 U = - 2

1 0 0

and U AU = diag -1, 1, 3 .

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