Using Eigenvectors to produce a Diagonal matrix

tomeatworld
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Homework Statement


If A=[{5,3},{-2,-2}], find the eigenvectors of A. Using these eigenvectors as matrix P, find P-1 and thus prove P-1AP is diagonal.


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The Attempt at a Solution


So i can get the eigenvectors to be <3,-1> and <1,-2> corresponding to eigenvalues 4 and -1 respecitively. The problem however, is choosing which vector should be the first column of the matrix P. I used <3,-1> as the first column, and didn't find a diagonal matrix. Should I have? if not, how should I choose which is the first row? I don't mind trying one then the other while revising, but if it's three 3x3 matricies and I'm in a exam, trying all posiilities isn't really an option. How should you choose?
 
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A= P D P-1

Where P is matrix of eigenvectors and D is matrix of eigenvalues on a diagonal


First you get the eigenvalues, then you get the eigenvectors. Your eigenvectors are wrong. Recheck your work to verify that eigenvectors are <-3,1> and <-1,2>.

Once you do that you can set P=[-3 -1; 1 2], D=[4 0; 0 -1], P-1=[-0.4 -0.2; 0.2 0.6] Do all the work to verify these results.

PDP-1 = [5 3; -2 -2]
 
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tomeatworld's eigenvectors are correct. It doesn't matter which vector you choose to be the first column of P, but how you choose will affect how the eigenvalues appear in the diagonal matrix D.
 
Mark44 said:
tomeatworld's eigenvectors are correct. It doesn't matter which vector you choose to be the first column of P, but how you choose will affect how the eigenvalues appear in the diagonal matrix D.

Ah you right, either eigenvector (+/-)[-3;1] and (+/-)[1;-2] would do for lambda=4 and lambda=-1, respectively
 
Emphasis mine:
tomeatworld said:
So i can get the eigenvectors to be <3,-1> and <1,-2> corresponding to eigenvalues 4 and -1 respecitively. The problem however, is choosing which vector should be the first column of the matrix P. I used <3,-1> as the first column, and didn't find a diagonal matrix. Should I have? if not, how should I choose which is the first row?
So, which did you do, make 3,-1 be the first row or the first column? Was your result

P^{-1}AP = \bmatrix 8.8 &amp; -5.6 \\ 8.4 &amp; -5.4\endbmatrix

If so, you constructed your P matrix as

P=\bmatrix 3 &amp; -1 \\ 1 &amp; -2\endbmatrix

It should be

P=\bmatrix \phantom{-}3 &amp; \phantom{-1}1 \\ -1 &amp; -2\endbmatrix

The reason is that eigenvectors are column vectors. You computed them via

A\vec x = \lambda \vec x

Written that way, the eigenvectors of an n×n matrix have to be n×1 vectors: column vectors.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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