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# Lecture 34.pdf

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University of Waterloo

Mathematics

MATH 136

Robert Sproule

Winter

Description

Monday, March 31 − Lecture 34 : More on eigenvectors.
Concepts:
1. Recognize that λ is a1 eigenvalue of Aif and only if 1/λ is an ei1envalue of A . -1
2. Define algebraic and geometric multiplicity of an eigenvalue.
34.1 Proposition − Let A be an invertible matrix. If λ is a1 eigenvalue of A then 1/λ is an 1
eigenvalue of A . Furthermore, x is an1eigenvector associated to 1/λ if and on1y if x is 1
an eigenvector of λ . 1ence, E 1/λ1= E λ1
Proof : Suppose matrix A is invertible. Then A does not have an eigenvalue equal to zero
since (A – 0I)x = Ax = 0 only has the trivial solution (since A is invertible).
λ 1is a eigenvalue of A.
⇔ the matrix equation Ax = λ x has1a non-trivial solution x 1
⇔ Ax = 1 x f1 1some non-zero eigenvector
−1 −1
⇔ A Ax = A 1 x , 1 1
−1
⇔ x =1A λ x . 1 1
⇔ A x = (1/λ ) x .
1 1 1
⇔ A x = (1/λ )x h1s a non-trivial solution
⇔ 1/λ i1 an eigenvalue ofA with x as an e1genvector.
Hence, E 1/λ1= E λ1
34.2 Definitions
Suppose λ is1an eigenvalue of the matrix A and E denotesλ1ts associated eigenspace.
The dimension of E is λ1lled the geometric multiplicity of λ . 1
Recall that if λ is an eigenvalue of the matrix Aand det(A − λ I) = (λ + 1) = 0. we say
1 1 1
that the algebraic multiplicity of the eigenvalue λ 1 is k.
Caution – One may suspect that the algebraic multiplicity of an eigenvalue λ must be 1
equal to its geometric multiplicity. But we will see that this is not always the case.
34.2.1 Example − Let A be the matrix, The determinant of A is 2 and so A is invertible. Compare the eigenvalues and
eigenvectors A with the eigenvalues and eigenvectors of A . State the algebraic and
geometric multiplicities of the eigenvalues for both matrices.
Solution :
To find the eigenvectors associated to 1 we solve for (A − 1I)x = 0:
and get x = α(1, 0, 0) + β(0, 1, 0). Thus the basis of E ha1 two vectors. Since the
dimension of E 1 is two, 1 is an eigenvalue of geometric multiplicity 2.
We also easily determine in the same way that the basis for E is {(2,0,1)} .
−1
We easily construct A and obtain:
Proceeding as above we obtain λ 1,2= 1 and λ =3½, as predicted by the preceding
theorem.
We also obtain as the eigenspace associated to λ 1, 2= 1,
E 1 α(1, 0, 0) + β(0, 1, 0) = span{(1, 0, 0) , (0, 1, 0)}
and eigenspace associated to λ 1/2 = 1,
E 1/2= span{(0,0,1)}
Note that A and A have the same eigenspaces span{(1, 0, 0) , (0, 1, 0)} and
span{(0,0,1)}. Remark – The reader may have noticed that the situation encountered when
determining the basis of Null(A – λ I) isibit different from what we usually see
when solving for systems Ax = b. So it is a good idea to practice this well.
34.2.2 Example − Verify that the matrix,
has characteristic equation det(A − λI) = ( λ + 1) = 0 3
Hence the algebraic multiplicity of the eigenvalue λ = −1 i1 3.
When the eigenvectors are calculated by solving (A − (−1)I)x = 0 we obtain as a
solution set x = α(0, −3, 1) + β(1, 0, 0) .
So E =1span{(0, −3, 1) , β(1, 0, 0)}. We see that the dimension of the eigenspace E is 1
2. It’s geometric multiplicity 2 while its algebraic multiplicity is 3.
34.2.4 Remarks − Suppose A is a 3 × 3 matrix. Then the matrix A may have
− only one eigenvalue of algebraic multiplicity 3,
− only 1 eigenvalue of algebraic multiplicity 1 and 1 of algebraic multiplicity 2
− three eigenvalues each of algebraic multiplicity 1.
Furthermore, A may have
− only one eigenspace

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