# Linear Algebra - Normal Operators

## Homework Statement

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

## 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(Tk) is the set of v in V such that Tkv = 0. To prove the first part, I want to show that Tkv = 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
Homework Helper
Gold Member
Ker(T) is the set of v in V such that Tv = 0. Ker(Tk) is the set of v in V such that Tkv = 0. To prove the first part, I want to show that Tkv = 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?

Let Tk(v) = Tk-1(T(v)) = 0. By the induction hypothesis, since Tk-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
Homework Helper
Gold Member
oops, disregard my hint, I made a mistake!

oops, disregard my hint, I made a mistake!

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

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
Homework Helper
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 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
Homework Helper
??? 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?

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(Tk).

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

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

Dick
Homework Helper
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)?

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!

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
Homework Helper
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.

How would you diagonalize:

$$\left( \begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} \right)$$

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

Dick
Homework Helper
How would you diagonalize:

$$\left( \begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} \right)$$

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*).

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

Thanks!