Lecture 9 eigenvalues - 5-1 & 5-2
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Transcript of Lecture 9 eigenvalues - 5-1 & 5-2
5
5.1
© 2012 Pearson Education, Inc.
Eigenvalues and Eigenvectors
EIGENVECTORS AND
EIGENVALUES
Slide 5.1- 2© 2012 Pearson Education, Inc.
EIGENVECTORS AND EIGENVALUES
Definition: An eigenvector of an nxn matrix A is a
nonzero vector x such that Ax=λx for some scalar
λ.
A scalar λ is called an eigenvalue of A if there is a
nontrivial solution x of Ax=λx;
such an x is called an eigenvector corresponding to λ.
Eigenvectors may NOT be 0
Eigenvalue may be 0
Eigenvectors - Example
Are u and v eigenvectors of A, for:
𝐴 =1 65 2
, 𝐮 =6−5
, 𝑣 =3−2
Slide 5.1- 3© 2012 Pearson Education, Inc.
Finding Eigenvectors - Example
Show that 7 is an eigenvalue of A from
previous example and find corresponding
eigenvectors.
Slide 5.1- 4© 2012 Pearson Education, Inc.
Eigenspace
General solution of Ax=λx is same as Nul
space of (A – λI)
Subspace of Rn
Called “Eigenspace”
Contains 0 and all eigenvectors
Slide 5.1- 5© 2012 Pearson Education, Inc.
Finding Basis for Eigenspace - Example
A has eigenvalue of 2. Find basis for
corresponding eigenspace.
𝐴 =4 −1 62 1 62 −1 8
Slide 5.1- 6© 2012 Pearson Education, Inc.
Slide 5.1- 7© 2012 Pearson Education, Inc.
Geometric Interpretation
The eigenspace, shown in the following figure, is
a two-dimensional subspace of R3.
Slide 5.1- 8© 2012 Pearson Education, Inc.
EIGENVALUES of Triangular Matrix
Theorem 1: The eigenvalues of a triangular
matrix are the entries on its main diagonal.
Proof: For simplicity, consider the 3x3 case.
If A is upper triangular, the (A – λI) has the form
11 12 13
22 23
33
11 12 13
22 23
33
λ 0 0
λ 0 0 λ 0
0 0 0 0 λ
λ
0 λ
0 0 λ
a a a
A I a a
a
a a a
a a
a
Slide 5.1- 9© 2012 Pearson Education, Inc.
EIGENVALUES of Triangular Matrix
The scalar λ is an eigenvalue of A if and only if the equation (A – λI) = 0 has a nontrivial solution, i.e., iff the equation has a free variable.
Because of the zero entries in (A – λI), it is easy to see that (A – λI) = 0 has a free variable if and only if at least one of the entries on the diagonal of A – λI is zero.
This happens if and only if λ equals one of the entries a11, a22, a33 in A.
Slide 5.1- 10© 2012 Pearson Education, Inc.
EIGENVECTORS – Linear Independence
Theorem 2: If v1, …, vr are eigenvectors that
correspond to distinct eigenvalues λ1, …, λr of an
nxn matrix A, then the set {v1, …, vr} is linearly
independent.
0 as Eigenvalue
If 0 is an eigenvalue:
Ax=0x=0 has non-trivial solution
A is NOT invertible
Additions to IMT:
(s) 0 is not an eigenvalue of A
(t) |A| ≠ 0
Slide 5.1- 11© 2012 Pearson Education, Inc.
5
5.1
© 2012 Pearson Education, Inc.
Eigenvalues and Eigenvectors
THE CHARACTERISTIC
EQUATION
Finding Eigenvalues - Example
Find the eigenvalues of 𝐴 =2 33 −6
Slide 5.1- 13© 2012 Pearson Education, Inc.
Characteristic Equation
characteristic equation: |A – λI| = 0
Scalar equation
Degree n, where A is nxn
Solutions are eigenvalues of A
Slide 5.1- 14© 2012 Pearson Education, Inc.
Slide 5.2- 15© 2012 Pearson Education, Inc.
THE CHARACTERISTIC EQUATION
Example 2: Find the characteristic equation of
5 2 6 1
0 3 8 0
0 0 5 4
0 0 0 1
A
Slide 5.2- 16© 2012 Pearson Education, Inc.
SIMILARITY
If A and B are nxn matrices, then A is similar to
B if there is an invertible matrix P such that
P-1AP = B, or, equivalently
A = PBP-1
Changing A into B is called a similarity
transformation.
Slide 5.2- 17© 2012 Pearson Education, Inc.
SIMILARITY
Theorem 4: If nxn matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).