Hermitian operators – a few theorems

References: edX online course MIT 8.05.1x Week 4.

Sheldon Axler (2015), Linear Algebra Done Right, 3rd edition, Springer. Chapter 7.

A hermitian operator {T} satisfies {T=T^{\dagger}}. [Axler (and most mathematicians, probably) refers to a hermitian operator as self-adjoint and uses the notation {T^*} for {T^{\dagger}}.]

As preparation for discussing hermitian operators, we need the following theorem.

Theorem 1 If {T} is a linear operator in a complex vector space {V}, then if {\left\langle v,Tv\right\rangle =0} for all {v\in V}, then {T=0}.

Proof: The idea is to show something even more general, namely that {\left\langle u,Tv\right\rangle =0} for all {u,v\in V}. If we can do this, then setting {u=Tv} means that {\left\langle Tv,Tv\right\rangle =0} for all {v\in V}, which in turn implies that {Tv=0} for all {v\in V}, implying further that {T=0}.

Zwiebach goes through a few stages in developing the proof, but the end result is that we can write

\displaystyle   \left\langle u,Tv\right\rangle \displaystyle  = \displaystyle  \frac{1}{4}\left[\left\langle u+v,T\left(u+v\right)\right\rangle -\left\langle u-v,T\left(u-v\right)\right\rangle \right]+\ \ \ \ \ (1)
\displaystyle  \displaystyle  \displaystyle  \frac{1}{4i}\left[\left\langle u+iv,T\left(u+iv\right)\right\rangle -\left\langle u-iv,T\left(u-iv\right)\right\rangle \right] \ \ \ \ \ (2)

Note that all the terms on the RHS are of the form {\left\langle x,Tx\right\rangle } for some {x}. Thus if we require {\left\langle x,Tx\right\rangle =0} for all {x\in V}, then all four terms are separately 0, meaning that {\left\langle u,Tv\right\rangle =0} as desired, completing the proof. \Box

Although we’ve used the imaginary number {i} in this proof, we might wonder if it really does restrict the result to complex vector spaces. That is, is there some other decomposition of {\left\langle u,Tv\right\rangle } that doesn’t required complex numbers that would still work?

In fact, we don’t need to worry about this, since there is a simple counter-example to the theorem if we consider a real vector space. In 2-d or 3-d space, an operator {T} that rotates a vector through {\frac{\pi}{2}} always produces a vector orthogonal to the original, resulting in {\left\langle v,Tv\right\rangle =0} for all {v}. In this case, {T\ne0} so the theorem is definitely not true for real vector spaces.

Now we can turn to a few theorems about hermitian operators. First, since every operator on a finite-dimensional complex vector space has at least one eigenvalue, we know that every hermitian operator has at least one eigenvalue. This leads to the first theorem on hermitian operators.

Theorem 2 All eigenvalues of hermitian operators are real.

Proof: Since at least one eigenvalue {\lambda} exists, let {v} be the corresponding non-zero eigenvector, so that {Tv=\lambda v}. We have

\displaystyle  \left\langle v,Tv\right\rangle =\left\langle v,\lambda v\right\rangle =\lambda\left\langle v,v\right\rangle \ \ \ \ \ (3)

Since {T=T^{\dagger}} we also have

\displaystyle  \left\langle v,Tv\right\rangle =\left\langle T^{\dagger}v,v\right\rangle =\left\langle Tv,v\right\rangle =\left\langle \lambda v,v\right\rangle =\lambda^*\left\langle v,v\right\rangle \ \ \ \ \ (4)

Equating the last two equations, and remembering that {\left\langle v,v\right\rangle \ne0}, we have {\lambda=\lambda^*}, so {\lambda} is real. \Box

Next, a theorem on the eigenvectors of distinct eigenvalues.

Theorem 3 Eigenvectors associated with different eigenvalues of a hermitian operator are orthogonal.

Proof: Suppose {\lambda_{1}\ne\lambda_{2}} are two eigenvalues of {T}, and {v_{1}} and {v_{2}} are the corresponding eigenvectors. Then {Tv_{1}=\lambda_{1}v_{1}} and {Tv_{2}=\lambda_{2}v_{2}}. Taking an inner product, we have

\displaystyle   \left\langle v_{2},Tv_{1}\right\rangle \displaystyle  = \displaystyle  \lambda_{1}\left\langle v_{2},v_{1}\right\rangle \ \ \ \ \ (5)
\displaystyle  \left\langle v_{2},Tv_{1}\right\rangle \displaystyle  = \displaystyle  \left\langle Tv_{2},v_{1}\right\rangle \ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \lambda_{2}\left\langle v_{2},v_{1}\right\rangle \ \ \ \ \ (7)

where in the last line we used the fact that {\lambda_{2}} is real when taking it outside the inner product. Equating the first and last lines and using {\lambda_{1}\ne\lambda_{2}}, we see that {\left\langle v_{2},v_{1}\right\rangle =0} as required.\Box

3 thoughts on “Hermitian operators – a few theorems

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