# Diagonalization help

1. May 9, 2010

### xicor

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

$$A=\left[\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & -1\\ 0 & 0 & 2\end{array}$$

a) Find the eigenvalues and corresponding eigenvectors of matrix A.
b)Find the matrix P that diagonalizes A.
c)Find the diagonal matrix D suh that A = PDP-1, and verify the equality.
d) Find the orthogonal matrix P that diagonalizes A.
e) Compute A4

2. Relevant equations

A = PDP-1,
AP = DP
A-I$$\lambda$$ = 0

3. The attempt at a solution

First I started by finding the eigenvalues values where $$\lambda$$=1 multipity two, 2. After this I tried finding the eigenvectors that form P and got v1=[0,-1,1] from $$\lambda$$=2 , and {v2, v3} = {[0, 1, 0], [0, 0, 1]}. From this I constructed the P matrix and got $$P=\left[\begin{array}{ccc}0 & 0 & 0\\ -1 & 1 & 0\\ 1 & 0 & 1\end{array}$$ and $$D=\left[\begin{array}{ccc}2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}$$ and this is where I get confused. The P matrix doesn't work in the form AP = PD and you can't find the inverse of P since the top row is all zeros. Once I figure this out, parts d and e should be straight-forward. Can someone point me to where I'm making a mistake here please. Thanks to everybody who helps.

2. May 9, 2010

### jbunniii

$v_3 = [0, 0, 1]$ is not an eigenvector corresponding to $\lambda = 1$. Just perform the multiplication to see that $A v_3 \neq v_3$.

3. May 9, 2010

### vela

Staff Emeritus
Show us how you found the eigenvectors because only one actually is correct.

4. May 9, 2010

### jbunniii

Both $v_1$ and $v_2$ are correct, actually.

5. May 9, 2010

### vela

Staff Emeritus
Apparently, I can't add. :)

6. May 9, 2010

### xicor

Alright, then there is something I'm possibly missing about eigenvectors. The first eigenvector was found by plugging $$\lambda$$ = 2 into the (A - I$$\lambda$$) matrix producing $$\begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & 0\end{bmatrix}$$ which gives the equations x1 = 0 and x2 = -x3 and constructing the vector from x3 gives x3[0, -1, 1] where the first eigenvector v1 = [0, -1, 1]. I then used the other eigenvalue $$\lambda$$ = 1 and found the matrix $$\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 1\end{bmatrix}$$ and the way I'm interpreting the book's reasoning for dealing with these matrixes, the eigenvectors that form would be x2[0, 1, 0] + x3[0, 0, 1] which are also the eigenvectors. Clearly there is something different I need to do when dealing with this kind of matrix.

7. May 9, 2010

### vela

Staff Emeritus
The matrix you got for the eigenvalue equal to 1 gives you the equation x3=0. It doesn't tell you anything about the other components. Do you see what the other eigenvector should be now?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook