Diagonalization & Eigen vectors proofs

[Bertrandkis]

Homework Statement

Question 1:
A) Show that if A is diagonalizable then $$A^{T}$$ is also diagonalizable.

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

Homework Statement

Question 2
If A and B are Similar matrices, then show that $$A^{2}$$ and $$B^{2}$$
are similar

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

Homework Statement

Question 3
Every matrix A is Similar itself

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?

 

[HallsofIvy]

Homework Statement

Question 1:
A) Show that if A is diagonalizable then $$A^{T}$$ is also diagonalizable.

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

Looks good.

Homework Statement

Question 2
If A and B are Similar matrices, then show that $$A^{2}$$ and $$B^{2}$$
are similar

The Attempt at a Solution

If A and B are similar then $$P^{-1}AP$$ = $$B$$

We know that $$P^{-1}A^{k}P$$ =$$D^{k}$$

You only know that if P is diagonallizable. That is not assumed in this problem.

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

Looks like "overkill" to me. If you are asked only about A² and B², why use a fact about the k^th power? If P^-1AP= B, then
B²= (P^-1AP)(P^-1AP)= (P^-1A)(PP^-1)(AP).

Homework Statement

Question 3
Every matrix A is Similar itself

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?

How about just taking P= I?

 

[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?

 

[HallsofIvy]

Yes, that works nicely.

 

[mathboy]

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

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