Non-diagonalisable matrix of a linear trans.

[pibeta]

Homework Statement

If the matrix of a linear transformation relative to a basis is non-diagonalisable, then for any other choice of basis, it too will be non-diagonalisable. Prove this is the case.

Homework Equations

Similar matrices.

The Attempt at a Solution

Let A_PB=[T]_BP_B, where A is a nxn matrix.

But I'm stuck. Not sure what to do next.

 

[Office_Shredder]

What is your definition of diagonalizable and non-diagonalizable? It makes a pretty big difference since there are a couple ways to approach this depending on what we can assume and what we want to prove

 

[Dick]

If it's diagonalizable then it has a basis of consisting of n eigenvectors. The choice of the basis to represent the matrix doesn't change that.

 

[keniwas]

There might be an easier way to show this, but in physics diagonalizing matricies always leads to eigenvalue problems... I believe your statement implies the matrix doesn't have a closed form for its eigenvalue equation. So
|A-\omega I|=0 wouldn't have n roots and hence can't be diagonalized.

But as office-shredder says, there is more than one way to diagonalize a matrix...

 

[Dick]

What is your definition of diagonalizable and non-diagonalizable? It makes a pretty big difference since there are a couple ways to approach this depending on what we can assume and what we want to prove

If you mean real diagonalizable vs complex diagonalizable, I'm not sure that's what this is about.

 

[Office_Shredder]

If you mean real diagonalizable vs complex diagonalizable, I'm not sure that's what this is about.

No, I mean, for example, the definition of non-diagonalizable that I would use, and a lot of other people, is something like "a matrix which is not diagonal under any change of basis". Making the whole question kind of moot. Someone thought this question was worth asking, so obviously there's a different definition floating around.

I would assume that a diagonalisable matrix would have a diagonal and the off-diagonal entries are all zero, and then the non-diagonalisable matrix would have off-diagonal entries not all zero - that's how I would define it (and I'm hoping that's the way my teacher wanted us to define it!)

That would be a diagonal and a... non-diagonal matrix. What is a diagonalizable matrix, and what is a non-diagonalizable matrix? You can't expect to solve the problem without knowing what you're talking about

 

[Dick]

No, I mean, for example, the definition of non-diagonalizable that I would use, and a lot of other people, is something like "a matrix which is not diagonal under any change of basis". Making the whole question kind of moot. Someone thought this question was worth asking, so obviously there's a different definition floating around.

Now you are just confusing me. What does being diagonalizable have to do with the choice of a basis? The problem just says show it's independent of basis. If you state the condition in terms of eigenvectors, it's clearly basis independent.

 

[pibeta]

Looks like I've caused too much trouble here. Thanks for trying.

Pibeta.

 

[Dick]

Looks like I've caused too much trouble here. Thanks for trying.

Pibeta.

Don't leave! Just show that the quality of a matrix being diagonalizable is independent of the basis by stating it in a way that is independent of the basis. I'm not sure what this flurry of confusion was about, but it's nothing about you causing trouble.

 

[HallsofIvy]

There might be an easier way to show this, but in physics diagonalizing matricies always leads to eigenvalue problems... I believe your statement implies the matrix doesn't have a closed form for its eigenvalue equation.

"Doesn't have a closed form for its eigenvalue equation"? What does that mean?

So
|A-\omega I|=0 wouldn't have n roots and hence can't be diagonalized.

That doesn't follow. If an operator (on an n dimensional vector space) has n distinct roots then it is diagonalizable, but if not it still may be diagonalizable. As Dick said, it depends on the number of independent eigenvectors, not eigenvalues.

But as office-shredder says, there is more than one way to diagonalize a matrix...

No, office-shredder did not say that. He said there is more than one way to define "diagonalizable".

 

[pibeta]

Thank you for all of this help (I know what diagonalisable means, but didn't realize they was more than one way to define it). And I've decided to come back too.

Anyways, a hint I was given was to use [T]_C=P_C<-B[T]_B(P_C<-B)^-1 and prove the contradiction, which is the matrix is diagonalisable.

To show it is diagonalisable, I got (using the hint) [T]_C=P_BAP_B^-1 (where A=P_B[T]_B(P_B)^-1 )but I'm not sure if this is right and where to go from here...

 

[HallsofIvy]

Thank you for all of this help (I know what diagonalisable means, but didn't realize they was more than one way to define it). And I've decided to come back too.

Anyways, a hint I was given was to use [T]_C=P_C<-B[T]_B(P_C<-B)^-1 and prove the contradiction, which is the matrix is diagonalisable.

To show it is diagonalisable, I got (using the hint) [T]_C=P_BAP_B^-1 (where A=P_B[T]_B(P_B)^-1 )


but I'm not sure if this is right and where to go from here...

I'm glad to know that you know what "diagonalizable" means- that helps a lot! But to help you we still need to know what your definition of "diagonalizable" is! Please tell us.