Linear Algebra Proof, similar diagonalizable matrices

  • Thread starter stihl29
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  • #1
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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?
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
Dick
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You didn't really plug your expression for A into the expression for B, like you said you were going to. Now did you?
 
  • #6
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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.
 
  • #7
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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 dont see how either of them simplify does [P][Q]^-1 = I ?
 
  • #8
Dick
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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?
 
  • #9
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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.
 
  • #10
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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 ?
 
  • #11
Dick
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It is if you are clear on what Z is in terms of P and W, and why W^(-1)P and P^(-1)W are inverses.
 
  • #12
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yea i looked up the the reasons, thanks a ton guys.
 

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