- #1

skyturnred

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## Homework Statement

Find a matrix P that satisfies D=P[itex]^{-1}[/itex]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([itex]\lambda[/itex]I-D)=0

so ([itex]\lambda[/itex]-1)([itex]\lambda[/itex]-1)([itex]\lambda[/itex]-5)=0

so e-values are 1, 1, 5

So I found e-vector of A using [itex]\lambda[/itex]=1

I did this by solving for vector x in: ([itex]\lambda[/itex]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 [itex]\lambda[/itex]=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[itex]_{A}[/itex]P=P[itex]_{D}[/itex] 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[itex]^{-1}[/itex]AP, the matrix P above isn't even invertible. Where did I go wrong? Thanks!