Can someone please check this work (eigenvalues)

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The discussion centers on finding an invertible matrix S and a diagonal matrix D for the given matrix A, ensuring that S^(-1)AS=D. The original matrices provided were incorrect, particularly the eigenvalues in D, which should reflect the eigenvalues derived from the characteristic polynomial. After clarification, it was confirmed that the diagonal of D must contain the correct eigenvalues, which were identified as 6.5 and 2.5. The participant acknowledged a mistake in calculating the inverse of S and adjusting the order of eigenvectors to match the eigenvalues. Ultimately, the revised matrices were deemed correct, with reassurance that multiple valid S matrices could diagonalize A.
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Homework Statement


Let

<br /> A = \left[ \begin{array}{cc} -6 &amp; 0.25 \\ 7 &amp; -3 \end{array} \right]<br />

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



My answer is:

<br /> S = \left[ \begin{array}{cc} 1 &amp; -1 \\ 14 &amp; 2 \end{array} \right]<br />

<br /> D = \left[ \begin{array}{cc} -40 &amp; 0 \\ 0 &amp; -104 \end{array} \right]<br />

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 ><
 
Last edited:
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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...

<br /> <br /> \left[ \begin{array}{cc} 2 &amp; 1 \\ -14 &amp; 1 \end{array} \right] <br /> \left[ \begin{array}{cc} -6 &amp; .25 \\ 7 &amp; -3 \end{array} \right]<br /> \left[ \begin{array}{cc} 1 &amp; -1 \\ 14 &amp; 2 \end{array} \right] <br /> <br />

Which gave me my D which I guess is obviously wrong. <br /> <br /> D = \left[ \begin{array}{cc} -40 &amp; 0 \\ 0 &amp; -104 \end{array} \right]<br /> <br />

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.
 
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.
 
Last edited:
Oh dear, I did forget about that! Thanks! I'll try again.
 
Okay so,

<br /> <br /> S = \left[ \begin{array}{cc} 1 &amp; -1 \\ 14 &amp; 2 \end{array} \right]<br /> <br />

<br /> <br /> S = \left[ \begin{array}{cc} -2.5 &amp; 0 \\ 0 &amp; -6.5 \end{array} \right]<br /> <br />

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?
 
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.
 
Well, you'll have to trust that the homework scanner will accept any correct answer. (You could also multiply S by 2 and multiply (inverse S) by 1/2.) If not, you have good grounds for appeal.
 
  • #10
Yeah the answer is correct now. All this new material made me forget the earlier stuff :(

Thanks a lot, appreciate the help.
 

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