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Mathematics

MATH 4389

Almus

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Math 2331
Section 5.1 – Eigenvectors and Eigenvalues
1 5
Example: Let A =
0 2
v =
w ==1
−1
u = 1
Definition: An eigenvector of an n××n matrix A is a nonzero vector x such
that
Ax == λx for some scalar λ λ .
A scalar λ is an eigenvalue of A if there exists a nontrivial solution x of
Ax == λx ; such an x is called an eigenvector corresponding to λλ .
= λ λλ
1 4 2
Example: A = = 1 3
Is u = an eigenvector?
1
Is v== an eigenvector?
6 3
Example: A =
2 1
Show that 7 is an eigenvalue of A.
2 ▯ ▯
Notice that Ax = λx for ▯ .
x ≠ 0⇒ Ax − λIx = 0⇒ (A− λI)x = 0
▯ ▯
That is,x is an eigenvector corresponding to λ if and only if x∈Nul(A−λI) .
▯
Definition: Ifx is an eigenvector corresponding to λ , Nul(A− − λI)
is called the eigenspace forλ .
Notation: Eλ .
KEY FACT: λ is an eigenvalue of A if and only ifdet(A−λI) = 0 .
Reason:
3 Definition: Suppose A is an nxn matrix.
th
The characteristic polynomial of A is an n degree polynomial in λ :
p(λ) = det(A−λI)
4 2
Example: Let A = . Find all e-values, e-vectors, e-spaces, characteristic
1 3
polynomial.
4 FACT: The special solutions to (A−λI)x = 0 form a basis for Eλ . This basis consist
of e-vectors corresponding to e-value λ .
1 1
Example: Let A = . Find all e-values, e-vectors, e-spaces, characteristic
1 1
polynomial.
5 0 1
Example: Let A = 1 0 . Find all e-values, e-vectors, e-spaces, characteristic
polynomial.
6 1 −1 0
Example: Let A= −1 2 −1 . Find all e-values, e-vectors, e-spaces,
0 −1 1
characteristic polynomial.
8 2 1 0
Example: Let A= 02 5 . Find all e-values, e-vectors, e-spaces, characteristic
0 0 2
polynomial.
9 2 1 0
Example: Let A= 0 2 0 . Find all e-values, e-vectors, e-spaces, characteristic
0 0 2
polynomial.
Definition: Algebraic multiplicity of an eigenvalueλ is the degree of the
corresponding factor in the characteristic polynomial.
Geometric multiplicity of an eigenvalue is the dimension of the corresponding
eigenspace.
GM ≤ ≤ AM
Fact:
10 IMPORTANT FACTS:
1) An nxn matrix A has at most n distinct eigenvalues.
2) If A is a triangular matrix, then the eigenvalues of A are the diagonal entries.
2 1 −1
A= 5 4 ; 2, 5, 6.
0 0 6
3) The product of the eigenvalues of a matrix A equals det(A).
4) The sum of the eigenvalues of a matrix A equals trace(A) .
Here, trace(A) = a11 a 22..+ a nn the sum of the diagonal entries.
Example:
4 2
A =
1 3
Prdocut of the evalues: 10
Sum of the eigenvalues: 7 ; 2 ad 5.
11 Example: A is a 2x2 matrix with determinant -20 andtrace 1. What are the
eigenvalues of A?
5) 0 is an eigenvalue of A if and only if A is singular.
12 1
6) If A is invertible anλ ≠ 0, then λ is an eigenvalue of A if and only iλ is an
eigenvalue of A−1. The evectors for λ and 1 are the same.
λ
Note: Elimination changes eigenvalues!
13 Theorem: If{v ,v ,...}vare eigenvectors corresponding to distinct
1 2 r
eigenvaluesλ1λλ2,..λr, then the se{v1,2 ,..r}vis linearly independent.
Question: Ix is an eigenvector of A, whatA x = ?
14 Math 2331
Section 5.2 – The Characteristic Equation
A:n×××n matrix
The characteristic equation of Adet(A− λI)= 0
A scalarλ is an eigenvalue of A if and onlyλisatisfies the characteristic
equation det(A−−λI) = 0.
Characteristic polynomial of A p(λ)= det(A− λI)
Algebraic multiplicity of an eigenvλlis its multiplicity as a ropλ)o.
Example: The characteristic polynomial of a matrix is:
6 5 4
pλ) = λ −9λλ −10λλ
Find the eigenvalues and their algebraic multiplicities.
1 The Invertible Matrix Theorem (continued)
Let A be an n××n matrix. A is invertible if and only if
s. The number 0 is NOT an eigenvalue of A.
t. The determinant of A is not 0.
2 Similarity
Let A and B be two n× ×n matrices.
A is similar to B if there exists an invertible matrix P such that P AP = = B .
−1
This is equivalent to saying that A = = PBP .
−1
Similarity Transformation: A ֏ P AP
Theorem: If A and B are similar matrices, then they have the same
characteristic polynomial and the same eigenvalues (with the same
multiplicities).
3 Similar matrices have the SAME:
• Eigenvalues
• Determinant
• Trace
• Rank
• Number of linearly independent eigenvectors
• Jordan form (later)
DIFFERENT:
• 4 subspaces (row space, column space, etc.)
• Eigenvectors
FACT: If B = P AP and x▯is an eigenvector for A, then P x1▯ is an
=
eigenvector for B.
4 2 1 2 0
Warning: A = 0 2 and B = 0 2 have the same eigenvalues. However,
they are NOT similar!
Similar ▯ same eigenvalues
Same eigenvalues ▯ may be similar, may be not.
Example : A = 2 3 and B = 4 1 are not similar.
0 1 0 5
5 2 3
Example : A = 0 1 Find a matrix which is similar to A.
6 Math 2331
Section 5.3 – Diagonalization
Recall: Diagonal matrices --
1 Definition: A:n××n matrix
A can be diagonalized ( or “A is diagonalizable”) if A is similar to a
DIAGONAL matrix D .
That is, if there exists an invertible matrix P and a diagonal matrix D such
thatA = PDP −.
2 3 1 1 2 0 1 − −1
Example: =
0 5 0 1 0 5 0 1
2 Theorem: The Diagonalization Theorem
An n××n matrix A is diagonalizable if and only if A has n linearly
independent eigenvectors.
−1
In fact,A = PDP , with D a diagonal matrix, if and only if the columns of P
are n linearly independent vectors of A. In this case, the diagonal entries of D
are eigenvalues of A that correspond, respectively, to the eigenvectors in P.
P: The eigenvector matrix for A
D: The eigenvalue matrix for A.
In other words, A is diagonalizable if it has enough eigenvectors to form a
n n
basis forℝ . Such a basis is called an eigenvector basis oℝ .
3 4 2
Example: Diagonalize A =1 3.
−1
From Section 5.1: e-value 2, e-vector , e-value: 5, e-vector
1
4 1 −1 0
Example: Diagonalize A= − 2 −1
0 −1 1
From Section 5.1- e-values are 0, 1, 3.
−1 1
e-vector for 0: 1 e-vector for 1: 0 , e-vector for 3: −−2
1 1
5 FACTS:
1) A is diagonalizable if every eigenvalue has enough eigenvectors; GM=AM.
(That is, if the sum of the dimensions of eigenspaces equals n )
2) If an ×n matrix A has n distinct eigenvalues, then A is diagonalizable.
(Reason: the set of n correspond

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