The question:

Prove that if A ~ B then B ~ A

My attempt:

1.

2.

Sub 1 into 2:

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.

Printable View

- Jul 30th 2011, 06:13 AMGlitchAnother similar matrices Q
**The question:**

Prove that if A ~ B then B ~ A

**My attempt:**

1.

2.

Sub 1 into 2:

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 AMFernandoRevillaRe: Another similar matrices Q
- Jul 30th 2011, 12:02 PMobd2Re: Another similar matrices Q
Well you are given that there exists an invertible matrix such that

.

Can you do anything to this equality do get something contaning

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

- Jul 31st 2011, 02:47 AMHallsofIvyRe: Another similar matrices Q
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 to by taking .