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The Spectral Theorem in Complex and Real Inner Product Space

  1. Jan 6, 2013 #1

    I am going through Sheldon Axler - Linear Algebra Done right. The book States the Complex Spectral Theorem as :

    Suppose that V is a complex inner product space and T is in L(V,V). Then V has an orthonormal basis consisting of eigen vectors of T if and only if T is normal.

    The proof of this theorem seems fine. It uses the property that ||Tv|| = ||T*v|| for a normal operator T, where T* is the adjoint of T.

    However, the Real Spectral Theorem States that V has an orthonormal basis consisting of eigen vectors of T if and if only if T is self adjoint.

    My Doubt : Why does Real Spectral Theorem take into account only self adjoint operators as a necessary condition despite the fact that an operator can be normal and still not self adjoint. When it's normal, the property ||Tv|| = ||T*v|| should be still valid for real inner product space which leads to the desired result.

    Would be great if somebody could give me an insight. Thanks.
    Last edited by a moderator: Jan 6, 2013
  2. jcsd
  3. Jan 6, 2013 #2
    The complex spectral theorem states that exactly the normal operators are the operators which are orthogonally diagonalizable.
    The real spectral theorem states that it are the self-adjoint operators. Why the difference?

    The real spectral theorem asks for which matrices [itex]A\in M_n(\mathbb{R})[/itex], there exists an orthogonal basis [itex]\{v_1,...,v_n\}[/itex] and real numbers [itex]\lambda_i[/itex] such that [itex]Av_i=\lambda v_i[/itex].

    Of course, if A satisfies the real spectral theorem, then it satisfies the complex spectral theorem. But vice versa is not the case. Let A be a normal operator, then it satisfies the complex spectral theorem. This means that there exists an orthogonal basis [itex]\{v_1,...,v_n\}[/itex] and complex numbers [itex]\lambda_i[/itex] such that [itex]Av_i=\lambda v_i[/itex].

    But in order to satisfy the real spectral theorem, we demand the [itex]\lambda_i[/itex] to be real (and we demand that entries of A to be real).
    So the matrices satisfying the real spectral theorem are exactly the (real) normal matrices with real eigenvalues. Now, it turns out that if a normal matrix has only real eigenvalues, then it is self-adjoint. This is why only self-adjoint matrices satisfy the real spectral theorem.
  4. Jan 6, 2013 #3
    Hi Micromass,

    So, if i prove that if the eigen values of a normal operator T's matrix are all real, then T is self adjoint , this should prove the real spectral theorem from the complex spectral theorem.

    Attempt: Given that : T T* = T* T and T is real .
    To prove that : T is self adjoint.

    Proof : T is normal => ||Tv||=||T*v|| . Now, this means from the complex spectral theorem that T has a diagonal matrix with complex entries.

    But, T is real => while calculating the modulus of the column vectors, we can deduce that
    the entries on the diagonal are actually real with 0 imaginary components.

    => T is self adjoint since a self adjoint operator has all real eigen values.

    Thanks Micromass :)
  5. Jan 6, 2013 #4


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    Here is an excerpt from the math 4050 notes on my web page:

    Cor: Structure of normal operators.
    Assume T:V-->V is a normal operator on a finite dimensional inner product space.
    1) If V is a complex space, with minimal polynomial m(T) = ∏(t-cj)^dj, all cj distinct, then V decomposes into an orthogonal direct sum of eigenspaces Vj = ker(T-cj). I.e., all dj = 1, and there is an orthonormal basis of V in which the matrix of T is diagonal.

    2) If V is a real space, with minimal polynomial m = ∏(t-ci)^di∏qj^ej, all ci distinct real scalars, and all qj distinct irreducible real monic quadratic polynomials, then
    i) V is an orthogonal direct sum of the invariant subspaces ker∏(T-ci) and ker∏qj(T).
    ii) The eigenspaces ker∏(T-ci) decompose into one dimensional invariant subspaces, and the invariant subspaces ker∏qj(T) decompose into indecomposable invariant two dimensional subspaces, on each of which qj is the minimal polynomial.
    iii) In particular all di =1, and all ej =1. The matrix of T on the eigenspace ker∏(T-ci) is diagonal with ci along the diagonal, and the matrix of T on ker∏qj(T) is a block matrix with 2 by 2 matrices of form
    |aj -bj |
    |bj aj |, along the diagonal, where the roots of qj are aj ± i bj.
    We get all the theorems from steps 1) and 2) by induction.
  6. Jan 7, 2013 #5
    Hi Mathwonk, Thanks a lot. These are helpful.

    My another query was this lemma :

    Suppose T in L(V,V) is self adjoint. Then T has real eigen values.V is a real inner product space

    Proof : Let n = dim V and choose v in V with v ≠ 0. Then

    (v, Tv , ......... , T^n v ) cannot be linearly independent because V has dimension n and we have n+1 vectors. Thus, there exist real numbers ao , ......, an, not all 0 such that

    0 = aov + a1Tv+.........+anT^n v
    = c (T^2 +mT + nI ) ( T^2 + rT+sI)(T - k1I) ......... ( T - k2I)

    the above factorisation is such that m^2<4n and r^2<4s

    Now, we also know that since T is self adjoint each of the quadratic expressions above is invertible . Hence, the roots lie amongst the linear expressions.

    However, How can the linear expressions guarantee a real root ( Although i can give an another proof to validate that a self adjoint does have real eigen values but does this expression solely substantiate the cause of T having real eigen values ? ) . Only when the degree of the above equation n is odd , can we be sure that it would by force, have a real eigen value else, we can't be sure.
    I would love to hear your views about this.

    Last edited: Jan 7, 2013
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