# Using Eigenvectors to produce a Diagonal matrix

1. Jun 3, 2010

### tomeatworld

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
None

3. 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?

2. Jun 3, 2010

### cronxeh

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]

Last edited: Jun 3, 2010
3. Jun 3, 2010

### Staff: Mentor

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.

4. Jun 3, 2010

### cronxeh

Ah you right, either eigenvector (+/-)[-3;1] and (+/-)[1;-2] would do for lambda=4 and lambda=-1, respectively

5. Jun 3, 2010

### D H

Staff Emeritus
Emphasis mine:
So, which did you do, make 3,-1 be the first row or the first column? Was your result

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

If so, you constructed your P matrix as

$$P=\bmatrix 3 & -1 \\ 1 & -2\endbmatrix$$

It should be

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

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.