Solving for Matrix P that Satisfies D=P^{-1}AP

[skyturnred]

Homework Statement

Find a matrix P that satisfies D=P$^{-1}$AP (A and D are similar)

A=

2 2 -1
1 3 -1
-1 -2 2

D=

1 0 0
0 1 0
0 0 5

Homework Equations

The Attempt at a Solution

OK, so I know how to find a matrix P for A, but I DONT know how to find the specific P that gets the specific A.. anyways here is my work so far

Since A and D are similar, e-values of D are the same as those of A

It is easy to find the e-values of D --> det($\lambda$I-D)=0
so ($\lambda$-1)($\lambda$-1)($\lambda$-5)=0
so e-values are 1, 1, 5

So I found e-vector of A using $\lambda$=1

I did this by solving for vector x in: ($\lambda$I-A)x=0

I found the following vector: x=t[-2, 1, 0]+w[1, 0, 1] where t and w are elements of the reals

doing the same for $\lambda$=5 I get: x=t[-1 -1 1]

so a P for A (not necessarily the proper P) is

-2 1 -1
1 0 -1
0 1 1

Using the same procedure for D as for A above, I get a P for D to be

0 0 0
0 1 0
1 0 1

This is where I have no idea what to do. I remember vaguely reading somewhere that the P in question is the matrix that transforms the P for A from above to the P for D

so I solve

P$_{A}$P=P$_{D}$ and get the following as a matrix

0.25 0.50 0.25
0.75 0.50 0.75
0.25 -0.5 0.25

...but checking in D=P$^{-1}$AP, the matrix P above isn't even invertible. Where did I go wrong? Thanks!

 

[fzero]

-2 1 -1
1 0 -1
0 1 1

Try applying this to A. The matrix P is known as the matrix that diagonalizes A. There isn't much point in trying to find a different matrix to diagonalize D.