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Why this paradox in calculating eigen values for T*T ?

  1. Apr 2, 2013 #1
    Let T be an operator on the vector space V and let λ1, ..... , λn be it's eigen values including multiplicity .

    Lets find the eigen values for the operator T*T then ( where T* refers to the adjoint operator . <u,v> denotes inner product of u and v )

    < Tv , Tv > = < λv, λv >
    = λλ° < v , v >
    = |λ|2 < v , v>

    ( where λ° is the conjugate of λ . )

    => <T*T v, v> = < |λ|2 v, v >

    => T*T has an eigen value |λ|2 for the same eigen vector v which T possesses.

    This can be inferred because :

    Suppose f is a linear functional on V . Then there is a unique vector v in V such that f(u) =<u,v> for all u in V.

    ( A linear functional on V is a linear map from V to the scalars F , As sheldon axler pg 117 says it ) The above argument must imply then that : T*T v = |λ|2 v ( Doesn't this imply v is an eigen vector of T*T ? )

    now, lets consider a matrix $$M(T) = \begin{bmatrix}
    1 & 3 \\
    0 & 2
    \end{bmatrix}$$

    [ 3 1 ]T is clearly an eigen vector with eigen value = 2 .

    (T* T ) =\begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix} multiplied by \begin{bmatrix}
    1 & 3 \\
    0 & 2
    \end{bmatrix}$$

    = \begin{bmatrix}
    1 & 3 \\
    3 & 13
    \end{bmatrix}$$

    M(T* T ) upon multiplication with [ 3 1 ]T should produce a vector equal to

    4 [ 3 1 ]T

    however , \begin{bmatrix}
    1 & 3 \\
    3 & 13
    \end{bmatrix} multiplied by [ 3 1 ]T = [ 6 22 ]T .

    Can you please advise why this paradox exists ? Am i making a mistake somewhere ?
    Thanks
     
    Last edited: Apr 2, 2013
  2. jcsd
  3. Apr 2, 2013 #2

    AlephZero

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    I don't understand the bra-ket notation, but I think

    is wrong.

    The eigenvectors of a matrix and its (conjugate) transpose are different in general. Your proof seems to assume they are the same. If ##Tx = \lambda x##, then ##x^*T^* = \lambda^* x^*##, but that doesn't make ##x^*## a (right) eigenvector of ##T^*##. (It is a left eigenvector of ##T^*##, of course).
     
  4. Apr 2, 2013 #3

    Fredrik

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    There's no bra-ket notation there, just inner products, absolute values, and the definition of the adjoint operator.

    Why would ##\langle T^*Tv,v\rangle=\langle|\lambda|^2v,v\rangle## imply that ##T^*Tv=|\lambda|^2 v##?

    I see that you tried to prove this, but the argument doesn't seem to make sense. There appears to be something missing from it as well. f(u)= what? Did you mean f(u)=v for all u? That actually implies that v=0.

     
  5. Apr 2, 2013 #4

    Fredrik

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    Note that if we let w be any vector that's orthogonal to v, we have ##\langle T^*Tv,v\rangle =\langle T^*Tv+w,v\rangle##.
     
  6. Apr 2, 2013 #5
    Thanks. <u,v> as fredrik pointed refers to the inner product of u and v . I get your argument regarding the adjoint operation.


    Hi Fredrik,
    Thanks for the answer. i have corrected it in the main thread , f(u) = <u,v> for all u in V.

    I think the equation < Tx , y > = < z, y > => Tx = z if and only if < Tx , y > = < z, y > is valid for all y in V. ( w[itex]\in[/itex] V ) .

    Lets consider the more general case of < w , y > = < z , y > , then w = z if and only if this relation is valid for all y [itex]\in[/itex] V.

    Proof : Suppose there exists an another different quantity w≠z | < w , y > = < z , y >

    Then : < w - z, y > = 0 for all y [itex]\in[/itex] V .
    => when y = w - z , then :

    <w-z,w-z> = 0 => w = z .
     
  7. Apr 2, 2013 #6
    <T*T v, v> = < |λ|2v, v > -------- (1)

    => I guess , T*T has an eigen value |λ|2 for the same eigen vector v which T possesses , if and only if (1) is valid for ALL v [itex]\in[/itex] V .
     
  8. Apr 2, 2013 #7

    Fredrik

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    If it holds for all v, I think the correct conclusion is that T*T is equal to ##|\lambda|^2I## where I is the identity operator. But more importantly, as you know, you have only proved that the equality holds for a specific v.
     
  9. Apr 2, 2013 #8

    micromass

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    It is clearly not true for all ##v\in V## since you took a very special ##v##, namely an eigenvector.
     
  10. Apr 2, 2013 #9

    AlephZero

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    Fair comment, but (as an ex-mathematician turned engineer) I never use the < > notation for inner products and have to think at least twice what "adjoint operator" means :smile:

    But in any case the problem here is with the math, not with the notation!
     
  11. Apr 2, 2013 #10
    Got it. Didn't keep into consideration all v in V have to satisfy the condition . Thanks !
     
    Last edited: Apr 2, 2013
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