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

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• Oct 8th 2011, 11:38 AM
skybluesea2010
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?
• Oct 8th 2011, 12:36 PM
Drexel28
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Quote:

Originally Posted by skybluesea2010
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?

It depends how much you know.

It's easy for example to show that if $m_1,m_2,m_3$ are the minimal polynomials for $H\to H$, $W\to W$, and $M/W\to M/W$ respectively, then $m_2,m_3\mid m_1$. The conclusion pretty much follows then from the fact that a linear operator $T:V\to V$ where $V$ is a $F$-space is diagonalizable if and only if $m_T$ splits over $F$ into distinct linear factors.

You can use the same theorem for part two. Namely, since the characteristic polynomial $p_M$ splits into linear factors and its minimal polynomial $m_M$ divides $p_M$ it must split into linear factors.

What do you think for the last part?
• Oct 8th 2011, 02:19 PM
skybluesea2010
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?
• Oct 8th 2011, 03:41 PM
Drexel28
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.
• Oct 8th 2011, 04:12 PM
skybluesea2010
Re: prove: restriction of an operator to invariant subspace is diagolizable?
• Oct 8th 2011, 04:14 PM
skybluesea2010
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Diagonalizable means minimum polynomial is a product of distinct linear operators
• Oct 8th 2011, 04:18 PM
Drexel28
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Quote:

Originally Posted by skybluesea2010

Sure, if that's the way you want to go. There is a theorem that says a linear operator $T:V\to V$ is diagonalizable if and only if its minimal polynomial $m_T$ factors into distinct linear terms. So the idea to show, for example, that $S:W\to W$ is diagonalizable is this: since $m_T(T)(x)=0$ for all $x\in V$ we evidently have that $m_T(S)(x)=0$ for all $x\in W$. Thus, one has from first principles that $m_S\mid m_T$. But, since $m_T$ factors into distinct linear factors this implies that $m_S$ factors into distinct linear factors and so $S$ is diagonalizable. Similarly, if $S_\ast$ denotes the induced map $V/W\to V/W$ one can show that $m_T$ annihilates $S_\ast$ and so $m_{S_\ast}\mid m_T$ from where you can draw the same conclusion. Make sense?
• Oct 8th 2011, 04:28 PM
skybluesea2010
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Definitely makes sense. Thank you so much. Now, it is clear to me.
• Oct 11th 2011, 02:24 PM
skybluesea2010
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
• Oct 11th 2011, 02:44 PM
skybluesea2010
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
• Oct 11th 2011, 03:23 PM
Drexel28
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Quote:

Originally Posted by skybluesea2010
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

Because the induced map $V/W\to V/W$ is given by $v+w\mapsto T(v)+W$, right? So, if $T(v)=0$ for all $v\in V$ then evidently $T(v+W)=0+W$ for all $v+W\in V/W$.
• Oct 11th 2011, 04:12 PM
skybluesea2010
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Quote:

Originally Posted by Drexel28
Because the induced map $V/W\to V/W$ is given by $v+w\mapsto T(v)+W$, right? So, if $T(v)=0$ for all $v\in V$ then evidently $T(v+W)=0+W$ for all $v+W\in V/W$.

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?
• Oct 11th 2011, 04:13 PM
skybluesea2010
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?
• Oct 11th 2011, 04:38 PM
Drexel28
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 $T$ are invariant under any commuting morphism $S$.
• Oct 11th 2011, 04:41 PM
skybluesea2010
Re: prove: restriction of an operator to invariant subspace is diagolizable?
Right.
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