# Another similar matrices Q

• Jul 30th 2011, 06:13 AM
Glitch
Another similar matrices Q
The question:
Prove that if A ~ B then B ~ A

My attempt:
1. $A = S^{-1}BS$
2. $B = S^{-1}AS$

Sub 1 into 2:
$A = S^{-1}(S^{-1}AS)S$
Let S = I
Thus A = A, and similarly B = B. Which is true. I'm pretty sure this isn't the correct process. What should I be doing to prove this? Thanks.
• Jul 30th 2011, 07:12 AM
FernandoRevilla
Re: Another similar matrices Q
Quote:

Originally Posted by Glitch
My attempt:1. $A = S^{-1}BS$
2. $B = S^{-1}AS$

$A = S_1^{-1}BS_1,\; B = S_2^{-1}AS_2$ etc.
• Jul 30th 2011, 12:02 PM
obd2
Re: Another similar matrices Q
Well you are given that there exists an invertible matrix $S$ such that

$A = S^{-1}BS$.

Can you do anything to this equality do get $B=$ something contaning $A$

Hint: If $S$ is invertible then so is $S^{-1}$.
• Jul 30th 2011, 02:41 PM
FernandoRevilla
Re: Another similar matrices Q
Explanatory note: The equalities I wrote in answer #2 were only to point out Glitch's mistake. Of course the best way of proving the symmetric property of similarity is using

$A=S^{-1}BS\Rightarrow SA=BS\Rightarrow SAS^{-1}=B\Rightarrow B=(S^{-1})^{-1}AS^{-1}$
• Jul 31st 2011, 02:47 AM
HallsofIvy
Re: Another similar matrices Q
Quote:

Originally Posted by Glitch
The question:
Prove that if A ~ B then B ~ A

My attempt:
1. $A = S^{-1}BS$
2. $B = S^{-1}AS$

Sub 1 into 2:
$A = S^{-1}(S^{-1}AS)S$
Let S = I
Thus A = A, and similarly B = B. Which is true. I'm pretty sure this isn't the correct process. What should I be doing to prove this? Thanks.

Yes, this is invalid. You have essentially shown that if you assume what you want to prove, you arrive at a true statement. That is not a valid proof method.

(You will sometimes see what is called "synthetic proof" where you start from the thing you want to prove and show the hypotheses. That is valid only if every step is "invertible". Then you could use reverse the proof and go from the hypotheses to the conclusion. That is the real proof. In a "synthetic proof" that reverse should be so obvious you don't have to explicitely do it.)

As both Obd2 and FernandoRevilla has said, you can go from $A= S^{-1}BS$ to $B= SAS^{-1}= S_2^{-1}AS_2$ by taking $S_2= S^{-1}$.