Eigenvalues and Eigenvectors Note2

8/4/2019 Eigenvalues and Eigenvectors Note2

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Eigenvalues and Eigenvectors:

Additional Notes


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1 2 1

6 1 0

1 2 1

A

!

1 2 3

1 1 2

6 , 2 , 3 ,

13 1 2

C C C

! ! !

1 2 3

0 4 6

0 , 8 , 9 ,

0 4 6

AC AC AC

! ! !

1 1 2 2 3 30 , 4 , 3 ,AC C AC C AC C ! ! !

Example. Consider the matrix

Consider the three column matrices

We have

In other words, we have

0,4 and 3 are eigenvalues of A, C1,C2 and C3 are eigenvectors

Ac cP!


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1 1 2

6 2 3

13 1 2

P

!

1

7 0 71

27 24 98432 12 8

P

!

1

7 0 7 1 2 1 1 1 2 0 0 01

27 24 9 6 1 0 6 2 3 0 4 084

32 12 8 1 2 1 13 1 2 0 0 3

P AP

! !

1

0 0 0

0 4 0

0 0 3

P AP

!

Consider the matrix P for which the columns are C1, C

2, and C3, i.e.,

we have det(P) = 84. So this matrix is invertible. Easy calculations give

Next we evaluate the matrix P-1AP.



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10 0 00 4 0

0 0 3

P AP

!

1

0 0 0

0 4 00 0 3

A P P

!

1

0 0 0

0 4 0

0 0 3

for 1,2,...

n n

n

A P P

n

!

!


Using the matrix multiplication, we obtain

which implies that A is similar to a diagonal matrix. In particular, we have


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.AC C P!

0 0 0 (0 is a zero vector)A P! !

Definition. LetA be a square matrix. A non-zero vectorCis called an

eigenvector o A i and only i there exists a number (real or complex)

such that

I such a number exists, it is called an eigenvalue o A. The vectorC

is called eigenvector associated to the eigenvalue

.

Remark. The eigenvectorCmust be non-zero since we have

or any number.


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1 2 16 1 0

1 2 1

A !

1 1 2 2 3 30 , 4 , 3 , AC C AC C AC C ! ! !

1 2 3

1 1 2

6 , 2 , 3 ,

13 1 2

C C C

! ! !


We have seen that

where

So C1 is an eigenvector o A associated to the eigenvalue 0.

C2 is an eigenvector o A associated to the eigenvalue -4

while C3 is an eigenvector o A associated to the eigenvalue 3.


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AC C P!

( ) 0nA I C P !

nA IP

det( ) 0.n

A IP {

Computation ofEigenvalues

For a square matrixA o order n, the number is an eigenvalue

i and only i there exists a non-zero vectorCsuch that

Using the matrix multiplication properties, we obtain

This is a linear system or which the matrix coe icient is

Since the zero-vector is a solution and Cis not the zero vector, then we must have

We also know that this system has one solution i and only i the matrix coe icient is

invertible, i.e.

det( ) 0.n

A IP !


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1 2

2 0

A

! det( ) 0.

nA IP !

1 2

(1 )(0 ) 4 02 0

P

P PP

! !

24 0P P !

1 17 1 17, and

2 2P P ! !


The equation

translates into

which is equivalent to the quadratic equation

Solving this equation leads to (use quadratic ormula)

In other words, the matrixA has only two eigenvalues.


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In general, or a square matrixA o order n, the equation

det( ) 0.nA IP !will give the eigenvalues o A.

This equation is called the characteristic equation orcharacteristic polynomial o A.

It is a polynomial unction in o degree n. There ore this equation will not

have more than n roots or solutions.

So a square matrixA o order n will not have more than n eigenvalues.


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0 0 00 0 0

.0 0 0

0 0 0

a

bD

c

d

!

0 0 0

0 0 0det( ) ( )( )( )( ) 0

0 0 0

0 0 0

n

a

bD I a b c d

c

d

P

PP P P P P

P

P

! ! !

Example. Consider the diagonal matrix

Its characteristic polynomial is

So the eigenvalues o D are a, b, c, and d, i.e. the entries on the diagonal.


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Remark. Any square matrixA has the same eigenvalues as its transposeAT

det( ) det( ) det( )T Tn n n

A I A I A I P P P ! !

a bA

c d

!

2 ( ) 0a b

a d bc a d ad bcc d

PP P P P

P

! ! !

For any square matrix o order 2,A, where

the characteristic polynomial is given by the equation

2 ( ) det( ) 0tr A AP P !

The number (a+d) is called the trace o A (denoted tr(A)), andthe number (ad-bc) is the determinant o A.

So the characteristic polynomial o A can be rewritten as


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Theorem. LetA be a square matrix o order n. I is an eigenvalue o A, then:

-1

1. is an eigenvalue o , or 1, 2,...

12. I A is invertible, then is an eigenvalue o .

3. A is not invertible i and only i 0 is an eigenvalue o A.

4. I is any number, then is

m mA m

A

P

P

P

E P E

!

!

an eigenvalue o .

5. I and are similar, then they have the same characteristic

polynomial (which implies they also have the same eigenvalues).

nA I

A B

E


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Computation ofEigenvectors

or ( ) 0n

AX X A I XP P! !

nA IP

LetA be a square matrix o order n and one o its eigenvalues.

LetX be an eigenvector o A associated to . We must have

This is a linear system or which the matrix coe icient is

Since the zero-vector is a solution, the system is consistent.

Remark. Note that i Xis a vector which satis iesAX= X,

then the vectorY= c X( or any arbitrary numberc) satis ies the

same equation, i.e.AY- Y.

In other words, i we know thatXis an eigenvector, then cXis also

an eigenvector associated to the same eigenvalue.


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1 2 1

6 1 0

1 2 1

A

!

3det( ) 0.A IP !

1 2 1

6 1 0 0

1 2 1

P

P

P

!

6 1 1 2( 1 ) 0

1 2 6 1

P PP

P

!

( 4)( 3) 0P P P !


First we look or the eigenvalues o A. These are given by the characteristic equation

I we develop this determinant using the third column, we obtain

By algebraic manipulations, we get

which implies that the eigenvalues o A are 0, -4, and 3.


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EIGENVECTORSASSOCIATED WITHEIGENVALUES

? A0AX !

2 0

6 02 0

x y z

x y

x y z

!

! !

6

13

y x

z x

!

!

1. Case =0. : The associated eigenvectors are given by the linear system

which may be rewritten by

The third equation is identical to the irst. From the second equation,

we have y = 6x, so the irst equation reduces to 13x +z = 0.

So this system is equivalent to


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1

6 6

13 13

x x

X y x x

z x

! ! !

1

6 ,

13

X c

!

So the unknown vectorXis given by

There ore, any eigenvectorXo A associated to the eigenvalue 0 is given by

where c is an arbitrary number.


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3

4

or ( 4 ) 0

AX X

A I X

!

!

5 2 0

6 3 0

2 3 0

x y z

x y

x y z

!

!

!

3[ 4 0]A I

5 2 1 06 3 0 0

1 2 3 0

2. Case =-4: The associated eigenvectors are given by the linear system

which may be rewritten by

We use elementary operations to solve it.

First we consider the augmented matrix


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1 2 3 0

5 2 1 0

6 3 0 0

1 2 3 0

0 8 16 0

0 9 18 0

Then we use elementary row operations to reduce it to a upper-triangular orm.

First we interchange the irst row with the irst one to get

Next, we use the irst row to eliminate the 5 and 6 on the irst column.

We obtain


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1 2 3 0

0 1 2 0

0 1 2 0

1 2 3 0

0 1 2 0

0 0 0 0

I we cancel the 8 and 9 rom the second and third row, we obtain

Finally, we subtract the second row rom the third to get


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1

2 2

1

x c

X y c c

z c

! ! !

1

2

1

X c

!

Next, we set z = c. From the second row, we get

y = 2z= 2c. The irst row will imply x = -2y+3z= -c. Hence

There ore, any eigenvectorXo A associated to the eigenvalue -4 is given by

wherec

is an arb

itrary numb

er.


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Summary: Let A bea square matrix. Assume is an

eigenvalue of A. In order to find theassociatedeigenvectors,

wed

o the foll

owing steps:

1. Write down the associated linear system

( ) 0n

AX X

or A I X

P

P

!

!

2. Solve the system.

3. Rewrite the unknown vectorXas a linear combination

o known vectors.

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Eigenvalues and Eigenvectors Note2

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