# Diagonalization & Eigen vectors proofs

1. Jan 10, 2008

### Bertrandkis

1. The problem statement, all variables and given/known data
Question 1:
A) Show that if A is diagonalizable then $$A^{T}$$ is also diagonalizable.

3. The attempt at a solution
We know that $$A$$ is diagonalizable if it's similar to a diagonal matrix.
So
$$A$$=$$PDP^{-1}$$
$$A^{T}$$=$$(PDP^{-1})^{T}$$
which gives
$$A^{T}$$=$$(P^{-1})^{T}DP^{T}$$ as $$D=D^{T}$$
Hence $$A^{T}$$ is diagonalizable

1. The problem statement, all variables and given/known data
Question 2
If A and B are Similar matrices, then show that $$A^{2}$$ and $$B^{2}$$
are similar

3. The attempt at a solution
If A and B are similar then $$P^{-1}AP$$ = $$B$$

We know that $$P^{-1}A^{k}P$$ =$$D^{k}$$
let k=2 therefore
$$P^{-1}A^{2}P$$ =$$B^{2}$$
hence $$A^{2}$$ and $$B^{2}$$ are similar

1. The problem statement, all variables and given/known data
Question 3
Every matrix A is Similar itself

3. The attempt at a solution
If A and A are similar then $$P^{-1}AP$$ =$$A$$ ? this does not make sense to me.
Alternatively, do we have to show that A has the same eigenvalues as A? This is obvious, is this then the proof?

2. Jan 10, 2008

### HallsofIvy

Staff Emeritus
Looks good.

You only know that if P is diagonallizable. That is not assumed in this problem.
Looks like "overkill" to me. If you are asked only about A2 and B2, why use a fact about the kth power? If P-1AP= B, then
B2= (P-1AP)(P-1AP)= (P-1A)(PP-1)(AP).

How about just taking P= I?

Last edited: Jan 10, 2008
3. Jan 14, 2008

### Bertrandkis

Thanks for the reply, I see where I went wrong.
I tried to use the method in question 2 and extend it to prove that :
IF A and B are similar matrices then $$A^{k}$$ and $$B^{k}$$ are similar for any non negative integer k.

This is what I got:
$$B^{k}$$=$$(P^{-1}AP)$$ $$(P^{-1}AP)$$ ......$$(P^{-1}AP)$$ (k times)
then Multiply the right hand side 2 elements at a time as u did we will end up with $$P^{-1}A^{k}P$$.
Is This the correct way to proove it?

4. Jan 14, 2008

### HallsofIvy

Staff Emeritus
Yes, that works nicely.

5. Jan 14, 2008

### mathboy

Not quite complete yet. Just write down that transpose of inverse is inverse of transpose.