# Diagonalization & Eigen vectors proofs

by Bertrandkis
Tags: diagonalization, eigen, proofs, vectors
 P: 25 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?
Math
Emeritus
Thanks
PF Gold
P: 39,502
 Quote by 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
Looks good.

 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 $$P^{-1}A^{k}P$$ =$$D^{k}$$
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 kth power? If P-1AP= B, then
B2= (P-1AP)(P-1AP)= (P-1A)(PP-1)(AP).

 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?
How about just taking P= I?
 P: 25 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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,502 Diagonalization & Eigen vectors proofs Yes, that works nicely.
P: 182
 Quote by Bertrandkis [b] 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
Not quite complete yet. Just write down that transpose of inverse is inverse of transpose.

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