prove: restriction of an operator to invariant subspace is diagolizable?

Hi all,

Can you guys help me out with solving the following problems.

Here is the problem 1 that I wanted to prove:

Let M: H--> H be diagonalizable (H vector space and W subspace of H). W is M- invariant. Prove that the restriction of M to W is daigonalizable and that T:M/W-->M/W is diagonalizable (note: M/W means M quotient W and not restriction of M to W, i.e the quotient space).

Problem 2:

Let M: H--> H has n distinct eigenvalues (note: dimension of H is n).

Let U: H-->H commute with M. show that U is diagonalizable?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Thanks a lot Drexel28

I am still not confused on how you figured out that m2, m3 divides m1, and how can we use that fact to answer our problem. Can you, please, elaborate more?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Quote:

Originally Posted by

**skybluesea2010** Thanks a lot Drexel28

I am still not confused on how you figured out that m2, m3 divides m1, and how can we use that fact to answer our problem. Can you, please, elaborate more?

Ok, let's start back a little bit then. How exactly do you define diagonalizable? If you try to work it out, we can work through it together when you get stuck.

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Can you please elaborate your last proof? Thanks

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Diagonalizable means minimum polynomial is a product of distinct linear operators

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Definitely makes sense. Thank you so much. Now, it is clear to me.

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Thanks for your help. Can you please tell me how de we formally prove that http://latex.codecogs.com/png.latex?m_T annihilates http://latex.codecogs.com/png.latex?S_%5Cast? Thanks a lot

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Thanks for your help. Can you please tell me how de we formally prove that annihilates ? Thanks a lot

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Quote:

Originally Posted by

**Drexel28** Because the induced map

is given by

, right? So, if

for all

then evidently

for all

.

have just one quick question: if you are given two linear operators which commute. One of which say F is diagonalizable. I proved that they have the same eigenvectors. How can I go from this to prove that the other linear operator say L is also diagonalisable?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

have just one quick question: if you are given two linear operators which commute. One of which say F is diagonalizable. I proved that they have the same eigenvectors. How can I go from this to prove that the other linear operator say L is also diagonalisable?

Re: prove: restriction of an operator to invariant subspace is diagolizable?

Quote:

Originally Posted by

**skybluesea2010** have just one quick question: if you are given two linear operators which commute. One of which say F is diagonalizable. I proved that they have the same eigenvectors. How can I go from this to prove that the other linear operator say L is also diagonalisable?

Ok, well, let's work this out together. So I'm sure you know that the eigenspaces of are invariant under any commuting morphism .

Re: prove: restriction of an operator to invariant subspace is diagolizable?