Diagonalization & Eigen vectors proofs


by Bertrandkis
Tags: diagonalization, eigen, proofs, vectors
Bertrandkis
Bertrandkis is offline
#1
Jan10-08, 06:22 AM
P: 25
1. The problem statement, all variables and given/known data
Question 1:
A) Show that if A is diagonalizable then [tex]A^{T}[/tex] is also diagonalizable.

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

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

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

We know that [tex]P^{-1}A^{k}P[/tex] =[tex]D^{k}[/tex]
let k=2 therefore
[tex]P^{-1}A^{2}P[/tex] =[tex]B^{2}[/tex]
hence [tex]A^{2}[/tex] and [tex]B^{2}[/tex] 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 [tex]P^{-1}AP[/tex] =[tex]A[/tex] ? 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?
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HallsofIvy
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#2
Jan10-08, 06:27 AM
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Quote Quote by Bertrandkis View Post
1. The problem statement, all variables and given/known data
Question 1:
A) Show that if A is diagonalizable then [tex]A^{T}[/tex] is also diagonalizable.

3. The attempt at a solution
We know that [tex]A[/tex] is diagonalizable if it's similar to a diagonal matrix.
So
[tex]A[/tex]=[tex]PDP^{-1}[/tex]
[tex]A^{T}[/tex]=[tex](PDP^{-1})^{T}[/tex]
which gives
[tex]A^{T}[/tex]=[tex](P^{-1})^{T}DP^{T}[/tex] as [tex]D=D^{T}[/tex]
Hence [tex]A^{T}[/tex] 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 [tex]A^{2}[/tex] and [tex]B^{2}[/tex]
are similar

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

We know that [tex]P^{-1}A^{k}P[/tex] =[tex]D^{k}[/tex]
You only know that if P is diagonallizable. That is not assumed in this problem.
let k=2 therefore
[tex]P^{-1}A^{2}P[/tex] =[tex]B^{2}[/tex]
hence [tex]A^{2}[/tex] and [tex]B^{2}[/tex] 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 [tex]P^{-1}AP[/tex] =[tex]A[/tex] ? 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
Bertrandkis is offline
#3
Jan14-08, 02:12 AM
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 [tex]A^{k}[/tex] and [tex]B^{k}[/tex] are similar for any non negative integer k.

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

HallsofIvy
HallsofIvy is offline
#4
Jan14-08, 05:22 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,902

Diagonalization & Eigen vectors proofs


Yes, that works nicely.
mathboy
mathboy is offline
#5
Jan14-08, 04:06 PM
P: 182
Quote Quote by Bertrandkis View Post
[b]
3. The attempt at a solution
We know that [tex]A[/tex] is diagonalizable if it's similar to a diagonal matrix.
So
[tex]A[/tex]=[tex]PDP^{-1}[/tex]
[tex]A^{T}[/tex]=[tex](PDP^{-1})^{T}[/tex]
which gives
[tex]A^{T}[/tex]=[tex](P^{-1})^{T}DP^{T}[/tex] as [tex]D=D^{T}[/tex]
Hence [tex]A^{T}[/tex] is diagonalizable
Not quite complete yet. Just write down that transpose of inverse is inverse of transpose.


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