# The Spectral Theorem in Complex and Real Inner Product Space

by vish_maths
Tags: complex, product, real, space, spectral, theorem
 P: 53 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.
 Mentor P: 18,244 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.
P: 53
 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 :)

 Sci Advisor HW Helper P: 9,470 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.
P: 53
 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.