# Homework Help: Diagonalization in R

1. Feb 1, 2016

### says

1. The problem statement, all variables and given/known data
Is the following matrix diagonalizable in R?

[ 2 1 0 ]
[ 1 3 -1 ]
[ -1 2 3 ]

2. Relevant equations

3. The attempt at a solution
I've checked my work and found one eigenvalue = 2, with the corresponding eigenvector = [1, 0, 1]. My question is -- Because I have one eigenvector, can I conclude that this matrix is diagonalizable in R? If I had two eigenvalues could I say it was diagonalizable in R and R2, or just R2?

2. Feb 1, 2016

### Ray Vickson

What are the other eigenvalues? There should be three eigenvalues altogether--including multiplicity--but they need not all be real.

And no, you could not say what you want (about diagonlizability in R or R2): it is either diagonalizable in R or not; if not, it may be diagonalizable in C (the complex field), or may not be diagonalizable at all, anywhere, if some of the eigenvalues are repeated. Note: repeated eigenvalues do not forbid diagonalizability, they just do not guarantee it.

3. Feb 1, 2016

### says

The question is only concerned with R and not C.

The matrix only gives 1 real eigenvalue and 1 real eigenvector. Can I conclude that the matrix is diagonalizable in R?

4. Feb 1, 2016

### Staff: Mentor

Is that what you meant to write? There should be three eigenvalues, not two, as you wrote above.
Is one of the eigenvalues repeated? As Ray already said, to determine whether a matrix is diagonalizable, you need to determine how many eigenvectors there are. If there are three of them, then the matrix is diagonalizable.

Last edited: Feb 1, 2016
5. Feb 1, 2016

### says

No, there is only one eigenvalue, not repeated. And only one eigenvector. The part of the question that says 'is the matrix diagonalizable in R?' made me wonder if it would be diagonalizable in R. I thought a nxn matrix was only diagonalizable if it has 3 distinct eigenvectors. I know one eigenvalue can have more than one eigenvector, but just the 'diagonalizable in R' made me think a bit...

6. Feb 1, 2016

### Staff: Mentor

@says, in the future, please post Linear Algebra questions in the Calculus & Beyond section. I moved this thread.

7. Feb 1, 2016

### says

Ok.

I just remembered D = P-1 A P , where P = eigenvectors (columns of P are the eigenvectors). The matrix A is not diagonalizable because P = only one eigenvector, so it has no inverse, hence, we can't solve the equation D = P-1 A P.

8. Feb 1, 2016

### Staff: Mentor

How many eigenvalues did you get? Your post was not clear on this.

9. Feb 1, 2016

### says

Only one real eigenvector. The question is only concerned with real eigenvectors.

10. Feb 1, 2016

### Staff: Mentor

OK, I buy that. There are three eigenvalues, but two of them are complex.
You can mark this thread as Solved, if you want.

11. Feb 1, 2016

### Ray Vickson

No, what you say is not true. There are THREE eigenvalues, but only one of them is real. There are three eigenvectors, but only one of them has all components real. The matrix is certainly diagonalizable in C. What you need to do is figure out if it is also diagonalizable in R.

12. Feb 1, 2016

### says

I have one eigenvector. Doesn't that mean it would be diagonalizable in R?

13. Feb 1, 2016

### Ray Vickson

Why would you think that? What do you know about matrices that are diagonalizable in R?

You keep saying you have one eigenvector. False, false, false! There are THREE eigenvectors. But, as I said before, only one of them is real!

Maybe your mis-statements are merely a matter of language or expression, but they could also indicate a basic lack of understanding. That is why I keep emphasizing the issue.

Note added in edit: sorry: several additional posts (including some that deal directly with the issue) did not appear on my screen at the time I composed this response. That type of mysterious delay has happened to me before, several times.

Last edited: Feb 1, 2016
14. Feb 2, 2016

### says

I just want to know if my conclusion is correct. I only have one real eigenvalue and one real eigenvector. The matrix isn't diagonalizable in R3, and I would say it's not diagonalizable in R because we need to have an nxn matrix with n eigenvectors for it to be diagonalizable.

15. Feb 2, 2016

### Ray Vickson

I have never, ever, heard the term "diagonalizable in R3". Where did you see it? What does it mean?

A matrix can be diagonalizable in R, or diagonalizable in C, or diagonalizable in some other scalar field K, but that is the extent of the term "diagonalizable".