Linear Algebra - Normal Operators

In summary, Dick thinks that the proof of the first part of the homework is trivial, but he is not sure if he knows how to prove the second part. He is asking for help.
  • #1
159
0

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.

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(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.
 
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  • #2
steelphantom said:
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 [tex]T^{k-1}(v)=0 \Rightarrow T(v)=0[/tex].Then write [tex]T^k(v)=T^{k-1}(T(v))[/tex]. 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?
 
  • #3
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.
 
  • #4
oops, disregard my hint, I made a mistake!
 
  • #5
quasar987 said:
oops, disregard my hint, I made a mistake!

That's ok! Thanks for your help. Does anybody else see a strategy for proving this?
 
  • #6
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.
 
  • #7
It's pretty trivial if you use that normal operators can be diagonalized by a unitary transformation. You don't want to use this?
 
  • #8
Dick said:
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?
 
  • #9
? 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?
 
  • #10
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
[tex]\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)
[/tex]

So Im(T) = Im(Tk). Is this the idea, or am I STILL not getting it? Thanks for your help, Dick.
 
  • #11
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)?
 
  • #12
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!
 
  • #13
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!
 
  • #14
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.
 
  • #15
How would you diagonalize:

[tex]

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

[/tex]

over the reals? The problem does not specify that V is a complex space.
 
  • #16
EBMath said:
How would you diagonalize:

[tex]

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

[/tex]

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*).
 
  • #17
Ah, I see...this implies <Tv,Tv>=0, a contradiction, right?

Thanks!
 

1. What is a normal operator in linear algebra?

A normal operator in linear algebra is a linear transformation that commutes with its adjoint. In other words, the adjoint and the operator have the same eigenvectors.

2. What is the significance of normal operators in linear algebra?

Normal operators have many important properties and applications in linear algebra. They play a crucial role in the spectral theorem, which states that any normal operator on a finite-dimensional complex vector space can be diagonalized by a unitary matrix. This allows for easier calculations and analysis of the operator's behavior.

3. How do you determine if an operator is normal?

To determine if an operator is normal, you can use the commutativity test. If the operator and its adjoint commute, then the operator is normal. Another way is to check if the operator's eigenvectors form an orthonormal basis, as this is a characteristic property of normal operators.

4. Can all operators be normal?

No, not all operators can be normal. In fact, a majority of operators are not normal. Only a special subset of operators, such as self-adjoint and unitary operators, can be classified as normal.

5. What are some real-world applications of normal operators?

Normal operators have a wide range of applications in fields such as quantum mechanics, signal processing, and image recognition. They are used to model physical systems and analyze their behavior, as well as to compress and enhance data in various forms of digital processing.

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