# Can someone please check this work (eigenvalues)

## Homework Statement

Let

$$A = \left[ \begin{array}{cc} -6 & 0.25 \\ 7 & -3 \end{array} \right]$$

Find an invertible S and a diagonal D such that $$S^{-1}AS=D$$

## Homework Equations

I basically have the question answered, just ONE problem.

## The Attempt at a Solution

$$S = \left[ \begin{array}{cc} 1 & -1 \\ 14 & 2 \end{array} \right]$$

$$D = \left[ \begin{array}{cc} -40 & 0 \\ 0 & -104 \end{array} \right]$$

I asked the professor what was wrong and he said "check the eigenvalues (main diagonal of D). The eigenvectors look ok, but their order has to match that of the eigenvalues.

I've checked over and over, the math works out(S^-1 A S = D) , but I can't see what's wrong with the order. It looks perfectly fine for me.

All I need to figure out is what the correct order is. Please help ><

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## Answers and Replies

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Dick
Homework Helper
The math does not work out 'fine'. S^(-1)AS=[[-5/2,0],[0,-13/2]] as I work it out. I'll agree with your professor. The eigenvalues are wrong, but the eigenvectors are right. How did you do that?

Well D is what I got from multiplying inverse of S, which was the eigen vectors I got multiplied by A, and then again by S, so...

$$\left[ \begin{array}{cc} 2 & 1 \\ -14 & 1 \end{array} \right] \left[ \begin{array}{cc} -6 & .25 \\ 7 & -3 \end{array} \right] \left[ \begin{array}{cc} 1 & -1 \\ 14 & 2 \end{array} \right]$$

Which gave me my D which I guess is obviously wrong.

$$D = \left[ \begin{array}{cc} -40 & 0 \\ 0 & -104 \end{array} \right]$$

My eigenvalues I got were 6.5 and 2.5 from the characteristic polynomial. Is that what's supposed to go into D's main diagonal or something? I thought I had to multiply it out like that.

Dick
Homework Helper
That helps. Your inverse of S is wrong. Try multiplying (inverse S)*S. You don't get the identity. Do you? Did you forget to divide by a determinant? And yes, the diagonal of D should be the eigenvalues.

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Oh dear, I did forget about that! Thanks! I'll try again.

Okay so,

$$S = \left[ \begin{array}{cc} 1 & -1 \\ 14 & 2 \end{array} \right]$$

$$S = \left[ \begin{array}{cc} -2.5 & 0 \\ 0 & -6.5 \end{array} \right]$$

Is this perfectly correct? I only get one more submission for this, so I need to make sure this is 100% correct. I'm only unsure because the prof said that "the eigenvectors look okay, but order has to match eigenvalues). Is that matched?

Dick
Homework Helper
You mean D= on the second line, right? And there's more than one S that will diagonalize A. So I'm not promising your homework system will accept it. But that looks ok to me. At least you have the eigenvalues on the diagonal.

Whoops yeah, I meant D, and yeah I found that there were other values of S that would diagonalize A, some of which were just switching the signs of the elements around. I think you're right, the answer should be correct now.

Dick