5 5.2 © 2012 pearson education, inc. eigenvalues and eigenvectors the characteristic equation
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© 2012 Pearson Education, Inc.
5
5.2
Eigenvalues and Eigenvectors
THE CHARACTERISTIC EQUATION
Slide 5.2- 2 © 2012 Pearson Education, Inc.
DETERMINANATS
Let A be an matrix, let U be any echelon form obtained from A by row replacements and row interchanges (without scaling), and let r be the number of such row interchanges.
Then the determinant of A, written as det A, is times the product of the diagonal entries u11, …, unn in U.
If A is invertible, then u11, …, unn are all pivots (because and the uii have not been scaled to 1’s).
n n
( 1)r
nA I
Slide 5.2- 3 © 2012 Pearson Education, Inc.
DETERMINANATS
Otherwise, at least unn is zero, and the product u11… unn is zero.
Thus
, when A is invertible when A is not invertible
product of( 1)
det pivots in
0,
r
A U
Slide 5.2- 4 © 2012 Pearson Education, Inc.
DETERMINANATS
Example 1: Compute det A for .
Solution: The following row reduction uses one row interchange:
1 5 0
2 4 1
0 2 0
A
1
1 5 0 1 5 0 1 5 0
0 6 1 0 2 0 0 2 0
0 2 0 0 6 1 0 0 1
A U
Slide 5.2- 5 © 2012 Pearson Education, Inc.
DETERMINANATS
So det A equals . The following alternative row reduction avoids the
row interchange and produces a different echelon form.
The last step adds times row 2 to row 3:
This time det A is , the same as before.
1( 1) (1)( 2)( 1) 2
1/ 3
2
1 5 0 1 5 0
0 6 1 0 6 1
0 2 0 0 0 1/ 3
A U
0( 1) (1)( 6)(1/ 3) 2
Slide 5.2- 6 © 2012 Pearson Education, Inc.
THE INVERTIBLE MATRIX THEOREM (CONTINUED) Theorem: Let A be an matrix. Then A is
invertible if and only if:
s. The number 0 is not an eigenvalue of A.
t. The determinant of A is not zero.
Theorem 3: Properties of Determinants Let A and B be matrices.
a. A is invertible if and only if det .
b. .
c. .
n n
n n0A
det (det )(det )AB A Bdet detTA A
Slide 5.2- 7 © 2012 Pearson Education, Inc.
PROPERTIES OF DETERMINANTS
d. If A is triangular, then det A is the product of the entries on the main diagonal of A.
e. A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar factor.
Slide 5.2- 8 © 2012 Pearson Education, Inc.
THE CHARACTERISTIC EQUATION
Theorem 3(a) shows how to determine when a matrix of the form is not invertible.
The scalar equation is called the characteristic equation of A.
A scalar λ is an eigenvalue of an matrix A if and only if λ satisfies the characteristic equation
λA I
det( λ ) 0A I
n n
det( λ ) 0A I
Slide 5.2- 9 © 2012 Pearson Education, Inc.
THE CHARACTERISTIC EQUATION
Example 2: Find the characteristic equation of
Solution: Form , and use Theorem 3(d):
5 2 6 1
0 3 8 0
0 0 5 4
0 0 0 1
A
λA I
Slide 5.2- 10 © 2012 Pearson Education, Inc.
THE CHARACTERISTIC EQUATION
The characteristic equation is
or
5 λ 2 6 1
0 3 λ 8 0det( λ ) det
0 0 5 λ 4
0 0 0 1 λ
(5 λ)(3 λ)(5 λ)(1 λ)
A I
2(5 λ) (3 λ)(1 λ) 0
2(λ 5) (λ 3)(λ 1) 0
Slide 5.2- 11 © 2012 Pearson Education, Inc.
THE CHARACTERISTIC EQUATION Expanding the product, we can also write
If A is an matrix, then is a polynomial of degree n called the characteristic polynomial of A.
The eigenvalue 5 in Example 2 is said to have multiplicity 2 because occurs two times as a factor of the characteristic polynomial.
In general, the (algebraic) multiplicity of an eigenvalue λ is its multiplicity as a root of the characteristic equation.
4 3 2λ 14λ 68λ 130λ 75 0
n n det( λ )A I
(λ 5)
Slide 5.2- 12 © 2012 Pearson Education, Inc.
SIMILARITY If A and B are matrices, then A is similar to B if
there is an invertible matrix P such that , or, equivalently, .
Writing Q for , we have .
So B is also similar to A, and we say simply that A and B are similar.
Changing A into is called a similarity transformation.
n n1P AP B
1A PBP
1P 1Q BQ A
1P AP
Slide 5.2- 13 © 2012 Pearson Education, Inc.
SIMILARITY Theorem 4: If matrices A and B are similar, then
they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).
Proof: If then,
Using the multiplicative property (b) in Theorem (3), we compute
----(1)
n n
1B P AP1 1 1 1λ λ ( λ ) ( λ )B I P AP P P P AP P P A I P
1
1
det( λ ) det ( λ )
det( ) det( λ ) det( )
B I P A I P
P A I P
Slide 5.2- 14 © 2012 Pearson Education, Inc.
SIMILARITY
Since , we see from equation (1) that .
Warnings:1. The matrices and
are not similar even though they have the same eigenvalues.
1 1det( ) det( ) det( ) det 1P P P P I det( λ ) det( λ )B I A I
2 1
0 2
1 0
0 2
Slide 5.2- 15 © 2012 Pearson Education, Inc.
SIMILARITY
2. Similarity is not the same as row equivalence. (If A is row equivalent to B, then for some invertible matrix E ). Row operations on a matrix usually change its eigenvalues.
B EA
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