Math 115 Lab 10 Solutions

1. Consider the two matrices

A :=

2 3 31 0 11 1 2

,B :=

0 1 03 0 12 0 0

.Calculate that these matrices have the same characteristic polynomial. Then show that one is diago-nalizable and the other is not.Answer: We can calculate directly that cA() = cB() = (+1)2( 2). But, the 1-eigenspace forA is the span of two basic eigenvectors, [1, 1, 0]T , [1, 0, 1]T , whereas for B the 1-eigenspace is only1-dimensional spanned by [1, 1, 2]T . Hence A is diagonalizable and B is not because we cannot finda basis for eigenvectors for B.

2. Consider the matrix,

A =[1 55 1

].

Show that A is orthogonally diagonalizable and find an orthogonal matrix P which diagonalizes A.Answer: A is symmetric, hence orthogonally diagonalizable. The characteristic polynomial of A canbe calculated to be ( 6)(+ 4). Hence, the eigenvalues are 6,4. Solving (A 6I)v = 0 we get thespan of [1, 1]T . Likewise, solving (A+ 4I)v = 0 we get the span of [1,1]T . The matrix

P =22

[1 11 1

]diagonalizes A. Clearly P is orthogonal.

3. Let P be an orthogonal matrix, and S a symmetric matrix. Show that P1SP is symmetric.Answer: The hypothesis means that PT = P1, and ST = S. Hence,

(P1SP )T = (PT )ST (P1)T = P1SP.

4. Show that for any square matrix A, AAT is orthogonally diagonalizable.Answer: AAT is symmetric, hence orthogonally diagonalizable because we have a criterion that saysthat a matrix is orthogonally diagonalizable if and only if it is symmetric.

5. Let T be a linear transformation.

(a) If T is defined by T (~x) = ~x, show that the standard matrix for T has eigenvalue = 1. ( We saythat T has eigenvalue = 1).

(b) Suppose T : R3 R3 takes a vector and projects it onto the plane given by the equation

2x y + 2z = 0.

i. List the eigenvalues of T .ii. For each eigenvalue determine the corresponding eigenspace and its dimension.iii. Is T diagonalizable? (That is, is the standard matrix for T a diagonalizable matrix?)

Hint: none of these questions require extensive computation.

Answer:

(a) Since T (~x) = A(~x) = 1~x), then = 1 is an eigenvalue of A.

(b) i. The eigenvalues are 0 and 1. The transformation fixes the plane, so any vector in the planehas eigenvalue 1 (ie. A(~x) = 1~x for any vector ~x in the plane. Any multiple of the normalvector to the plane is projected to the 0-vector, hence has eigenvalue 0, (ie. A~n = ~0 = 0~n.)

ii. E1 is the plane with equation 2x y + 2z = 0. The plane has dimension 2, so dim(E1) = 2.E0 is the line spanned by the normal vector [2,1, 2]T . The line through the normal vectorhas dimension 1 so dim(E0) = 1.

iii. The matrix corresponding to this transformation is diagonalizable because we can find a basisfor R3 consisting of eigenvectors any pair of non-colinear vectors in the plane along withany non-zero vector on the normal line will do.

6. Prove the following: If A is similar to B and B is similar to C, then A is similar to C.

Solution: Given that B = P1AP , for an invertible matrix P and C = Q1BQ for an invertiblematrix Q, we substitute the expression for B into the expression for C to obtain C = Q1P1APQ orC = (PQ)1A(PQ), where PQ is an invertible matrix. Thus A is similar to C.