Determining if a matrix is diagonalizable with explanation

LinAlgStudent

1. The problem statement, all variables and given/known data

Determine if this matrix is diagonalizable and explain why or why not.

[ 2 1 0 0 0 ]
[ 0 2 1 0 0 ]
[ 0 0 2 1 0 ]
[ 0 0 0 2 1 ]
[ 0 0 0 0 2 ]




2. Relevant equations

No equations provided and the use of a calculator will be prohibited on the test, so thats out of the question. I can find the determinant via expansion, but that doesn't help really.



3. The attempt at a solution

So I'm pretty confident that this is not diagonalizable because the only eigenvalue seems to be 2 with multiplicity 5, although i'm not sure that actually proves anything. Additionally, I'm fairly certain that because the study guide has this as a 5x5 matrix with calculator use prohibited, there must be a way, just from the shape, to determine whether it's diagonalizable. Does anyone have any tips?

Thank you so much for any help!

Muphrid

One trick is to just take the matrix you have and row reduce it until you have only diagonal entries left. If you can do that, you have a diagonalizable matrix.

LinAlgStudent

it row reduces to the identity matrix. Is that indicative of anything specific?

micromass

Got a reference for that?? I don't think what you say is true...

The best way to do this exercise is find the eigenvectors and see if they span the space.

LinAlgStudent

So I've found all eigenvalues to be equal to 2, the first eigenvector i think is:

[1]
[0]
[0]
[0]
[0]

and the other four all all 5x1 zero vectors.

But what does that mean in the context of the question?

Ray Vickson

The matrix (A) above is the matrix representation of a linear transformation T with respect to the basis e_1=(1,0,0,0,0),...,e_5 = (0,0,0,0,1). If a transformation (i.e., a matrix) is diagonalizable, its matrix becomes diagonal when re-expressed in some other basis f_1, f_2, f_3, f_4, f_5. Since two similar matrices have the same eigenvalues, the "diagonal" would need to be 5 times the identity matrix, and that means that the f_i would have to be eigenvectors of T (i.e., of the matrix A). In turn, that means that there must be 5 linearly independent eigenvectors. You can determine the eigenspace of A and check whether its dimensionality is 5.

RGV

LinAlgStudent

Provided that all the eigenvectors of A are zero vectors save for one of them which has only one nonzero entry, wouldn't that suggest that the eigenspace of A is only 1 dimensional? So, in order for A to be diagonalizable, the eigenspace would have to be of dimension 5, but it is of dimension 1 so it is not diagonalizable?

algebrat

For further exploration, research Jordan matrix and generalized eigenvector.

Yes, because of what I know, I see that it is in Jordan form, and so it is "as close to diagonal" as it will ever be.

I also like the method explained above, find the eigenvectors for the sole eigenvalue, "in the end there will be only one."

(The zero vector is never called an eigenvector, otherwise it would always be one, A0=λ0, though eigenvalues can be zero. )

Ray Vickson

Zero vectors do not count as eigenvectors (essentially by definition), so the eigenspace is, indeed, one-dimensional. So, YES: the matrix is not diagonalizable.

In fact, the matrix is already in its Jordan Canonical Form, and that consists of a single Jordan block of dimension 5. A diagonalizable matrix would have to have a diagonal Jordan Form.

RGV

Muphrid

Yes, I was mistaken; I thought I'd uncovered a quick way to work the problem.