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Linear Algebra - Normal Operators

  1. Apr 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that if T in L(V) is normal, then Ker(Tk) = Ker(T) and Im(Tk) = Im(T) for every positive integer k.

    2. Relevant equations

    3. 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.
     
  2. jcsd
  3. Apr 24, 2008 #2

    quasar987

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    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?
     
  4. Apr 24, 2008 #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.
     
  5. Apr 24, 2008 #4

    quasar987

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    oops, disregard my hint, I made a mistake!
     
  6. Apr 24, 2008 #5
    That's ok! Thanks for your help. Does anybody else see a strategy for proving this?
     
  7. Apr 24, 2008 #6

    Vid

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    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.
     
  8. Apr 24, 2008 #7

    Dick

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    It's pretty trivial if you use that normal operators can be diagonalized by a unitary transformation. You don't want to use this?
     
  9. Apr 24, 2008 #8
    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?
     
  10. Apr 24, 2008 #9

    Dick

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    ??? 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?
     
  11. Apr 24, 2008 #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.
     
  12. Apr 24, 2008 #11

    Dick

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    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)?
     
  13. May 21, 2008 #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!
     
  14. Jul 30, 2009 #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!
     
  15. Jul 30, 2009 #14

    Dick

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    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.
     
  16. Jul 30, 2009 #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.
     
  17. Jul 30, 2009 #16

    Dick

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    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*).
     
  18. Jul 30, 2009 #17
    Ah, I see...this implies <Tv,Tv>=0, a contradiction, right?

    Thanks!
     
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