Linear Algebra - Normal Operators

[steelphantom]

Homework Statement

Prove that if T in L(V) is normal, then Ker(T^k) = Ker(T) and Im(T^k) = Im(T) for every positive integer k.

Homework Equations

The Attempt at a Solution

Since T is normal, I know that TT* = T*T, and also that ||Tv|| = ||T*v|| and <Tv, Tv> = <T*v, T*v>.

Ker(T) is the set of v in V such that Tv = 0. Ker(T^k) is the set of v in V such that T^kv = 0. To prove the first part, I want to show that T^kv = 0 iff Tv = 0.

I'm not really sure where the fact that T is normal plays into this. I assume that proving the second part about Im(T) is done in much the same way. Thanks, as always, for any help.

 

[quasar987]

Ker(T) is the set of v in V such that Tv = 0. Ker(T^k) is the set of v in V such that T^kv = 0. To prove the first part, I want to show that T^kv = 0 iff Tv = 0.

Surely you can do the "<==" part, as this is true of any linear operator.

For the "==>" part, have you tried fiddling with induction? It's trivially true for k=1. Assume it's true for k-1. That is to say, assume T^{k-1}(v)=0 \Rightarrow T(v)=0.Then write T^k(v)=T^{k-1}(T(v)). How can you use that, the induction hypotheses, the fact that T is normal and the definition of the adjoint to get to Tv=0?

 

[steelphantom]

Let T^k(v) = T^k-1(T(v)) = 0. By the induction hypothesis, since T^k-1(T(v)) = 0, then T(T(v)) = T²(v) = 0. I'm not seeing where the normality of T and the adjoint will come into play here.

 

[quasar987]

oops, disregard my hint, I made a mistake!

 

[steelphantom]

oops, disregard my hint, I made a mistake!

That's ok! Thanks for your help. Does anybody else see a strategy for proving this?

 

[Vid]

T(v) = 0
Thus T^k(v) = T^k-1(T(v)) = T^k-1(0) = 0 since linear maps always map 0 to 0.

 

[Dick]

It's pretty trivial if you use that normal operators can be diagonalized by a unitary transformation. You don't want to use this?

 

[steelphantom]

It's pretty trivial if you use that normal operators can be diagonalized by a unitary transformation. You don't want to use this?

It's not that I don't want to; I just didn't think of it! I'm not sure if I know what you're saying.

So if T can be diagonalized, then M(T, B) is a diagonal matrix for some basis B. If M(T, B) = 0, then M(T, B)^k = 0. But if M(T, B)^k = 0, this doesn't necessarily mean that M(T, B) = 0. This isn't what you're talking about, is it?

 

[Dick]

? You don't want to show the matrix is zero. If M(T,B)=D where D is a diagonal matrix, describe ker(T) and Im(T) in terms of that basis. The matrix of T^k is D^k. Hence?

 

[steelphantom]

Ok, for the kernel, since M(T, B) is diagonal, M(T, B)*(x) = 0 = M(T, B)^k*(x), where (x) is a 1 x n vector consisting of all 0s. So Ker(T) = Ker(T^k).

For the image, since M(T, B) and M(T^k, B) are both diagonal matrices, when multiplied by a vector, they will produce another vector
\left(<br /> \begin{array}{ccc}<br /> e1 &amp; 0 &amp; 0\\<br /> 0 &amp; e2 &amp; 0\\<br /> 0 &amp; 0 &amp; en<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c}<br /> v1\\<br /> v2\\<br /> vn\\<br /> \end{array}<br /> \right)<br /> = \left(<br /> \begin{array}{c}<br /> e1*v1\\<br /> e2*v2\\<br /> en*vn\\<br /> \end{array}<br /> \right)<br />

So Im(T) = Im(T^k). Is this the idea, or am I STILL not getting it? Thanks for your help, Dick.

 

[Dick]

STILL not. Think of the diagonal basis {e1...en}. Pick an e_i. Consider the diagonal element D_ii. D_ii is either zero or non-zero. What might that have to do with e_i being in Ker(T) or Im(T)?

 

[harmlessthing]

So, you have that the matrix for T corresponding to an orthonormal basis of eigenvectors is the diagonal matrix
D = [a_1 0 ... 0
0 a_2
...
0 ... a_n]
where a_1, ... a_n are the eigenvectors corresponding to the basis of eigenvectors.

then the matrix corresponding to T^k is
D^k = [(a_1)^k 0
...
0 (a_n)^k]

correct? but i don't see how you can determine Null(T^k) = Null(T) and Range(T^k) = Range(T) with this... help please!

 

[EBMath]

Hm...can I bump this? I'm struggling with this same problem. First...this is a normal operator in an arbitrary inner product space, so I don't see how we know there is an orthonormal basis of eigenvectors. Also, I have a hunch that there is a less mechanical way to show this. Dimensionality?? But I just can't get my head around it.

Thanks!

 

[Dick]

Every normal operator can be diagonalized. That's the basic property you need. If you can think of a more basic proof that doesn't use this, more power to you.

 

[EBMath]

How would you diagonalize:

<br /> <br /> \left(<br /> \begin{array}{cc}<br /> 0 &amp; -1\\<br /> 1 &amp; 0<br /> \end{array}<br /> \right)<br /> <br />

over the reals? The problem does not specify that V is a complex space.

 

[Dick]

How would you diagonalize:

<br /> <br /> \left(<br /> \begin{array}{cc}<br /> 0 &amp; -1\\<br /> 1 &amp; 0<br /> \end{array}<br /> \right)<br /> <br />

over the reals? The problem does not specify that V is a complex space.

Ok. That's a good reason to find a more direct approach. You know ker(T) is a subspace of ker(T^2). Suppose ker(T^2) is not equal to ker(T). That means you have a v such that T(Tv)=0 and Tv is not 0. So Tv is in ker(T). Now use ker(T)=ker(T*).

 

[EBMath]

Ah, I see...this implies <Tv,Tv>=0, a contradiction, right?

Thanks!