Another similar matrices Q

**The question:**

Prove that if A ~ B then B ~ A

**My attempt:**

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

2. $\displaystyle B = S^{-1}AS$

Sub 1 into 2:

$\displaystyle A = S^{-1}(S^{-1}AS)S$

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.

Re: Another similar matrices Q

Quote:

Originally Posted by

**Glitch** **My attempt:**1.$\displaystyle A = S^{-1}BS$

2. $\displaystyle B = S^{-1}AS$

$\displaystyle A = S_1^{-1}BS_1,\; B = S_2^{-1}AS_2$ etc.

Re: Another similar matrices Q

Well you are given that there exists an invertible matrix $\displaystyle S$ such that

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

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

Hint: If $\displaystyle S$ is invertible then so is $\displaystyle S^{-1}$.

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

$\displaystyle A=S^{-1}BS\Rightarrow SA=BS\Rightarrow SAS^{-1}=B\Rightarrow B=(S^{-1})^{-1}AS^{-1}$

Re: Another similar matrices Q

Quote:

Originally Posted by

**Glitch** **The question:**

Prove that if A ~ B then B ~ A

**My attempt:**

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

2. $\displaystyle B = S^{-1}AS$

Sub 1 into 2:

$\displaystyle A = S^{-1}(S^{-1}AS)S$

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 $\displaystyle A= S^{-1}BS$ to $\displaystyle B= SAS^{-1}= S_2^{-1}AS_2$ by taking $\displaystyle S_2= S^{-1}$.