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?

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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 $\displaystyle m_1,m_2,m_3$ are the minimal polynomials for $\displaystyle H\to H$, $\displaystyle W\to W$, and $\displaystyle M/W\to M/W$ respectively, then $\displaystyle m_2,m_3\mid m_1$. The conclusion pretty much follows then from the fact that a linear operator $\displaystyle T:V\to V$ where $\displaystyle V$ is a $\displaystyle F$-space is diagonalizable if and only if $\displaystyle m_T$ splits over $\displaystyle F$ into distinct linear factors.

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

What do you think for the last part?

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?

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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?

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**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 $\displaystyle T:V\to V$ is diagonalizable if and only if its minimal polynomial $\displaystyle m_T$ factors into distinct linear terms. So the idea to show, for example, that $\displaystyle S:W\to W$ is diagonalizable is this: since $\displaystyle m_T(T)(x)=0$ for all $\displaystyle x\in V$ we evidently have that $\displaystyle m_T(S)(x)=0$ for all $\displaystyle x\in W$. Thus, one has from first principles that $\displaystyle m_S\mid m_T$. But, since $\displaystyle m_T$ factors into distinct linear factors this implies that $\displaystyle m_S$ factors into distinct linear factors and so $\displaystyle S$ is diagonalizable. Similarly, if $\displaystyle S_\ast$ denotes the induced map $\displaystyle V/W\to V/W$ one can show that $\displaystyle m_T$ annihilates $\displaystyle S_\ast$ and so $\displaystyle m_{S_\ast}\mid m_T$ from where you can draw the same conclusion. Make sense?

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?

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Originally Posted by

**skybluesea2010**

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

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

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

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?

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 $\displaystyle T$ are invariant under any commuting morphism $\displaystyle S$.

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