## The Spectral Theorem in Complex and Real Inner Product Space

Hi

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.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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 $A\in M_n(\mathbb{R})$, there exists an orthogonal basis $\{v_1,...,v_n\}$ and real numbers $\lambda_i$ such that $Av_i=\lambda v_i$. 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 $\{v_1,...,v_n\}$ and complex numbers $\lambda_i$ such that $Av_i=\lambda v_i$. But in order to satisfy the real spectral theorem, we demand the $\lambda_i$ 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.

 Quote by micromass 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 $A\in M_n(\mathbb{R})$, there exists an orthogonal basis $\{v_1,...,v_n\}$ and real numbers $\lambda_i$ such that $Av_i=\lambda v_i$. 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 $\{v_1,...,v_n\}$ and complex numbers $\lambda_i$ such that $Av_i=\lambda v_i$. But in order to satisfy the real spectral theorem, we demand the $\lambda_i$ 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.
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 :)

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## The Spectral Theorem in Complex and Real Inner Product Space

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.

 Quote by mathwonk 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.
Hi Mathwonk, Thanks a lot. These are helpful.

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.