Question

Linear Algebra: For what values of a (if any) is the matrix {{1,0,0},{1,a,0},{1,0,3}} defective? Explain your answer.

Answer 1

Matrix: \[\left[\begin{matrix}1 & 0 & 0 \\ 1 & a & 0 \\ 1 & 0 & 3\end{matrix}\right]\]

Answer 2

a = 0... than the first and second row are dependent

Answer 3

Hm, the correct answer is supposed to be a=1.

Answer 4

Right, sorry... I think you have to make a Jordan block out of the matrix, probably by subtractin row 2 of of row 3...

srry for that, I'm not too good with terminology ;P

Answer 5

haha it's okay. well i finally got it though.

Our nxn matrix A is diagonalizable if and only if there is a basis for R^n that consists of eigenvectors of A, which means we must have 3 vectors in our basis. We also see that if a is either 1 or 3, we will have double roots for our characteristic polynomial. That doesn't mean automatically though that a=1 or 3 will make the matrix defective. We have to check each case.

For a=1, our eigenvalues will be 1, 1, 3. When you find a basis for the eigenspace (i.e. the null space of A-eigenvalue*I), through solving the system A - I = 0, we find that there is only one free variable and thus only one eigenvector. Combine that with the eigenvector that results from eigenvalue 3 and we only have 2 vectors to make as a basis for R^3. This isn't enough, we're missing one more vector to complete the basis. But that's all we have, so a=1 makes the matrix defective.

For a=3, repeat the process and we get 2 free variables and thus two eigenvectors for the eigenvalue 3. Combine that with the eigenvector that results from eigenvalue 1, and we have 3 eigenvectors that we can use as a basis for R^3. Thus, a=3 will result in enough vectors that can make a basis for R^3 and so will not make the matrix defective.

Therefore a=1 is our only candidate that will make the matrix defective.