Thread: matrix algebra

Hi,

Consider the matrix A=

1 0 1
1 1 0
0 0 -1

a) find the characteristic polynomial of A
b) list the eigenvalues of A and their multiplicities
c) for the eigenvalues found in part B, describe the eigenvectors
d) is A diagonalizable?

My work:

a) We must find det(A-lambdaI) = so the characteristic polynomial is (1-lambda)(1-lambda)(-1-lambda) = (1-lambda)^2(-1-lambda)
b) The eigenvalues are 1,1,-1. 1 has an algebraic multiplicity of two and -1 has an algebraic multiplicity of one. What about the geometric multiplicity?
c) I'm not sure
d) No, it is not diagonalizable since the eigenvalues are not distinct (there is a duplicate)

Can someone please let me know where I went right and wrong and lead me in the right direction?

I agree with your answers to a) and b). As for c), you know that the eigenvalues are defined as constant values ( ) so that . The vectors for which this occurs are known as the eigenvectors. Since , if you substitute and each , you should be able to solve for the components of .

the answer to (b) is incomplete. the reason is, you have to answer (c) first.

i prefer to solve (A - λI)x = 0 myself (just a personal preference). the number of "free parameters" you have will give the dimension of the eigenspace (geometric multiplicity). this is always less than or equal to the algebraic multiplicity.

for (d), just because you have a repeated eigenvalue does NOT mean the matrix is not diagonalizable. for example, the identity matrix has only one eigenvalue, 1, but it's diagonal (and thus diagonalizable, just do nothing!). to answer (d) you HAVE to find the dimensions of the eigenspaces, so you have to finish (c). there's no short-cut.