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

**skybluesea2010** · October 8th 2011, 11:38 AM

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?

**Drexel28** · October 8th 2011, 12:36 PM

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?

**skybluesea2010** · October 8th 2011, 02:19 PM

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?

**Drexel28** · October 8th 2011, 03:41 PM

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.

**skybluesea2010** · October 8th 2011, 04:12 PM

Can you please elaborate your last proof? Thanks

**skybluesea2010** · October 8th 2011, 04:14 PM

Diagonalizable means minimum polynomial is a product of distinct linear operators

**Drexel28** · October 8th 2011, 04:18 PM

Originally Posted by skybluesea2010

Can you please elaborate your last proof? Thanks

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?

**skybluesea2010** · October 8th 2011, 04:28 PM

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

**skybluesea2010** · October 11th 2011, 02:24 PM

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

**skybluesea2010** · October 11th 2011, 02:44 PM

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

**Drexel28** · October 11th 2011, 03:23 PM

Originally Posted by skybluesea2010

Thanks for your help. Can you please tell me how de we formally prove that $m_T$ annihilates $S_\ast$ ? 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 $v+W\in V/W$ .

**skybluesea2010** · October 11th 2011, 04:12 PM

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?

**skybluesea2010** · October 11th 2011, 04:13 PM

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?

**Drexel28** · October 11th 2011, 04:38 PM

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 $S$ .

**skybluesea2010** · October 11th 2011, 04:41 PM

Right.

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