Diagonalizing Matrix A: Eigenvalues, Eigenvectors, Matrix P & D

[xicor]

Homework Statement

<br /> A=\left[\begin{array}{ccc}1 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; -1\\ 0 &amp; 0 &amp; 2\end{array}<br />

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

Homework Equations

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

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 <br /> P=\left[\begin{array}{ccc}0 &amp; 0 &amp; 0\\ -1 &amp; 1 &amp; 0\\ 1 &amp; 0 &amp; 1\end{array}<br /> and <br /> D=\left[\begin{array}{ccc}2 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 1\end{array}<br /> 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.

 

[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.

 

[vela]

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

 

[jbunniii]

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

Both v_1 and v_2 are correct, actually.

 

[vela]

Both v_1 and v_2 are correct, actually.

Apparently, I can't add. :)

 

[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 <br /> \begin{bmatrix}-1 &amp; 0 &amp; 0 \\ 0 &amp; -1 &amp; -1 \\ 0 &amp; 0 &amp; 0\end{bmatrix}<br /> 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 <br /> \begin{bmatrix}0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; -1 \\ 0 &amp; 0 &amp; 1\end{bmatrix}<br /> 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.

 

[vela]

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?