1. ## Induction matrix proof.

Apparently this is simple, but I'm not seeing it.

P, A are matrices, n is a real.

Use induction to prove that $(P^{-1}AP)^n = P^{-1}A^nP$.

Presumably the first step is:

$(P^{-1}AP)^n = P^{-n}A^nP^n$

I'm not sure what's next, or if that is correct.

2. Originally Posted by scorpion007
Apparently this is simple, but I'm not seeing it.

P, A are matrices, n is a real.

Use induction to prove that $(P^{-1}AP)^n = P^{-1}A^nP$.

Presumably the first step is:

$(P^{-1}AP)^n = P^{-n}A^nP^n$

I'm not sure what's next, or if that is correct.
In brief:

Step 1: True for n = 1.

Step 2: Assume true for n = k. That is, assume $(P^{-1}AP)^k = P^{-1} A^k P$ is true.

Step 3: Show true for n = k + 1. That is, assuming $(P^{-1}AP)^k = P^{-1} A^k P$ is true, show that $(P^{-1}AP)^{k+1} = P^{-1} A^{k+1} P$.

To do this, note that $(P^{-1}AP)^{k+1} = (P^{-1}AP)^{k} \cdot (P^{-1}AP)$ and substitute $(P^{-1}AP)^k = P^{-1} A^k P$.

3. Originally Posted by mr fantastic
In brief:

Step 1: True for n = 1.

Step 2: Assume true for n = k. That is, assume $(P^{-1}AP)^k = P A^k P$ is true.
Did you mean $P^{-1} A^k P$ in the RHS?

4. Originally Posted by scorpion007
Did you mean $P^{-1} A^k P$ in the RHS?
Yes. I have made the necessary edit.

5. Thanks, I see it now:

Expanding the last step would be:

$(P^{-1}AP)^{k+1} = (P^{-1}AP)^{k} \cdot (P^{-1}AP) =$
$(P^{-1} A^k P) \cdot (P^{-1}AP) = P^{-1}A^k\underbrace{PP^{-1}}_{I}AP=P^{-1}A^k\cdot AP=\color{red}P^{-1}A^{k+1}P$

Correct? Too verbose?

6. Originally Posted by scorpion007
Thanks, I see it now:

Expanding the last step would be:

$(P^{-1}AP)^{k+1} = (P^{-1}AP)^{k} \cdot (P^{-1}AP) =$
$(P^{-1} A^k P) \cdot (P^{-1}AP) = P^{-1}A^k\underbrace{PP^{-1}}_{I}AP=P^{-1}A^k\cdot AP=\color{red}P^{-1}A^{k+1}P$

Correct? Too verbose?
It's fine.

Now you need to write out the complete proof by induction, dotting the i's and crossing the t's.