Prove if B is similar to A then A is similar to B

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• Feb 10th 2010, 05:59 PM
mulaosmanovicben
Prove if B is similar to A then A is similar to B
I am having trouble with this proof.

A matrix B is similar to A if there exists a nonsingular matrix P such that

PAP^-1=B

The question is:

Prove if B is similar to A then A is similar to B

I have:

Assume B is similar to A. Then there exists P such that

PAP^-1=B.

multiplying both sides by P gives

PA=BP

then multiplying both sides by inverse of P gives

A=P^-1 *B*P

but I need to show A=PBP^-1. and I cannot figure it out. (Headbang) Thanks!!!!!!!
• Feb 10th 2010, 06:07 PM
tonio
Quote:

Originally Posted by mulaosmanovicben
I am having trouble with this proof.

A matrix B is similar to A if there exists a nonsingular matrix P such that

PAP^-1=B

The question is:

Prove if B is similar to A then A is similar to B

I have:

Assume B is similar to A. Then there exists P such that

PAP^-1=B.

multiplying both sides by P gives

PA=BP

then multiplying both sides by inverse of P gives

A=P^-1 *B*P

but I need to show A=PBP^-1. and I cannot figure it out. (Headbang) Thanks!!!!!!!

No, you do NOT need to prove \$\displaystyle A=PBP^{-1}\$ , which in general isn't true. What you need is to prove that there exists an invertible matrix \$\displaystyle Q\$ s.t. \$\displaystyle A=QBQ^{-1}\$...well, just choose \$\displaystyle Q=P^{-1}\$ (Wink)

Tonio
• Feb 10th 2010, 06:26 PM
mulaosmanovicben
Thanks Tonio! I actually skipped class the day the professor described invertible matricies (had to work on some homework that was due that day haha) but this solution sounds elegant and beautiful =D Thanks alot!!!!!!!!!!!!

EDIT: aah now I see invertible just means AB=BA=I