# Matrices, eigenvalues, invertibility

1. Apr 11, 2015

### blue4123

1. The problem statement, all variables and given/known data
For which $2x2$ matrices $A$ does there exist an invertible matrix
$S$ such that $AS=SD$, where
$D= \begin{bmatrix} 2 & 0\\ 0 & 3 \end{bmatrix}$?
Give your answers in terms of the eigenvalues of $A$.

2. Relevant equations
$A\lambda=\lambda\vec{v}$

3. The attempt at a solution
S is a $2x2$ matrix. If x and y are the 1st and second columns of A, then we can write the following:
Ax=2x
Ay=3y

So we can see from these two equations that 2 and 3 must be the corresponding eigenvalues of A and their respective eigenvectors must be linearly independent.

I was hoping someone could critique my solution attempt as I am not too confident that I fully answered the question or if there's a better way to express my answer

2. Apr 12, 2015

### BruceW

hi there, welcome to physicsforums :)
I think you meant to say that x and y are the first and second columns of S. I agree with what you're saying. But you haven't shown in what cases does S exist. The problem was to show when S exists. It will not exist for all choices of matrix A, but only a special subset.

Edit: ah, wait, sorry, the problem is to show for what kinds of matrix A, does S exist, such that D is diagonal with values 3,2. OK, I think you pretty much have the answer. You've shown that if S exists, A must have eigenvalues 3 and 2.

Last edited: Apr 12, 2015
3. Apr 12, 2015

### BruceW

yeah. sorry for my last post, now I have read your work through again, I think you have answered correctly. The problem is to find the set of all matrices A such that S exists and such that D is diagonal with values 2,3. So to enumerate this set, it is OK to assume S exists, and then find out what this implies for the matrix A. You found that A must have eigenvalues 2,3. So then the last bit of the problem would be to say that all matrices with eigenvalues 2,3 must be able to be diagonalized. And this is what you were implying when you said that the respective eigenvectors of A must be linearly independent.

So maybe you could write a little bit more detail about why if a matrix has eigenvalues 2,3, it must then have linearly independent eigenvectors. Apart from that, I think you have done everything you need for this problem.

4. Apr 12, 2015

### blue4123

Oh sorry, I did mean that x and y are the first and second columns of S, and not A.

5. Apr 12, 2015

### blue4123

That is the part I was a bit unsure of. I found that A must have eigenvalues 2,3. Are their corresponding eigenvectors always linearly independent?

6. Apr 12, 2015

### Zondrina

Perhaps a theorem may help:

Suppose $A \in M_{n \times n}(\mathbb{C})$. If $A$ has $n$ distinct eigenvalues, then any set of $n$ corresponding eigenvectors form a basis for $\mathbb{C}^n$.

Of course a basis is any nonsingular spanning set.

Think about what this means in terms of scalars. You can create any basis you desire, and all of the vectors inside of it will correspond to the eigenvalues, regardless of being scaled.