Homework Help: 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

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?