# Linear Algebra Proof, similar diagonalizable matrices

stihl29

## Homework Statement

Prove that if matrices A and b are similar and A is diagonalizable, then B is diagonalizable.

## Homework Equations

this shows that A and B are similar i believe
A = [P][P]^-1
and
D = [P]^-1 [A] [P]
means A is diagonalizable

## The Attempt at a Solution

I believe this is a simple proof but i want to know the reason behind each step of the proof.
this is what I've tried D =[p]^-1 ([P][P]^1)[P]
what i did was plug in A from the first equation into the second so i would get
[D]= ? this this right?

Homework Helper
D isn't equal to B unless B is already diagonal. The matrix 'P' that shows A is similar to B and the matrix 'P' that diagonalizes A don't have to be the same matrix. It is simple. You are just making it too simple.

VeeEight
You are right in plugging in your matrix A into the second equation, but you must use different matrices to conjugate A and B. Call one P and one Q.

stihl29
so, something like...
A = [P][P]^-1
B= [Q][A][Q]^-1

B =[P]^-1[A][P]
A =[Q]^-1[Q]
??
i'm pretty bad with proofs as you can tell.

Homework Helper
You didn't really plug your expression for A into the expression for B, like you said you were going to. Now did you?

VeeEight
Just take your two equations in your first post and in one of them (say, A = ...), switch the P's to Q's (since you don't know the same matrix works to conjugate both matrices) and plug it in the second formula, the same way you did. Then just mess around with the terms, you might have to use a formula about inverses.

stihl29
I was just seeing if what i did so far was correct,
so if i put
B =[P]^-1[A][P] into
A =[Q]^-1[Q]

A =[Q]^-1([P]^-1[A][P])[Q] ??

or

B = [Q]([P][P]^-1)[Q]^-1

i don't see how either of them simplify does [P][Q]^-1 = I ?

Homework Helper
Whatever happened to D? I thought you wanted to show B is diagonalizable? That means you want to show D=R^(-1)BR for some matrix R. What is R in terms of P and Q?

VeeEight
You are trying to show that B is diagonalizable, so you want your conclusion to be that D = Z-1BZ. So, plug A into the formula for D. Then use the formula for product of inverses.

stihl29
A = [P][P]^-1
and
D = [W]^-1 [A] [W]

D = [W]^-1 ([P][P]^-1) [W]
by matrix inverse property
D = Z-1 B Z

is that correct ?