So I want to eventually read the proof of the spectral theorem for compact self-adjoint linear operators in a Hilbert space but I keep forgetting the proof of the finite dimensional case I saw in a linear algebra course. I will thus write up this proof to hopefully remember and understand it better. First, I will lay the basic definitions and theorems for Hermitian spaces and then prove the big theorem in my next post. I will follow Artin’s Algebra.

First some definitions:

Definitions:

Hermitian form on a complex vector space $V$ is a map $V \times V \to \textbf{C},$ denoted by $\langle u,v \rangle,$ such that it is conjugate linear in the first argument, linear in the second, and Hermitian symmetric. Conjugate linear means that

$\langle \lambda u,v \rangle = \overline \lambda \langle u,v \rangle$

and Hermitian symmetric means that

$\langle u,v \rangle = \overline{\langle v,u \rangle}.$

A Hermitian form on $V$ is said to be positive definite if for all $u \in V,$ $\langle u,u \rangle \geq 0$ with equality iff $u = 0.$

A Hermitian space $(V,\langle \cdot, \cdot \rangle)$ is a complex vector space $V$ with a positive definite Hermitian form $\langle \cdot , \cdot \rangle.$ We will only work with finite dimensional vector spaces in this post. The standard Hermitian form on $\textbf{C}^n$ is

$\langle u,v \rangle = u^*v = \sum \overline{u_i}v_i.$

Let $\phi : V \to V$ be a linear operator on a Hermitian space $V,$ and let $M$ be the matrix of $\phi$ with respect to an orthonormal basis $B.$ The adjoint operator of $\phi,$ denoted $\phi^* : V \to V,$ is the operator whose matrix is the adjoint of $M,$ denoted $M^*$ (with respect to the same basis $B$). Recall that the adjoint of the matrix $[a_{ij}]$ is $[\overline{a_{ji}}],$ the transposed matrix with conjugated entries. Note that this is well defined : if we change $B$ to a new orthonormal basis $B'$ with a change of basis matrix $P,$ the new matrix of $\phi$ will be $\phi' = P^*MP.$ Therefore its adjoint will be

$\phi'^{*} = (P^*MP)^* = P^*M^*(P^*)^* = P^*M^*P$

which is just $M^*$ is the new basis, so the definition of the adjoint of an operator does not depend on the chosen orthonormal basis.

Finally, a normal operator is an operator that commutes with its adjoint, ie $\phi^* \circ \phi = \phi \circ \phi^*$ or equivalently $MM^* = M^*M$ for $M$ the matrix of $\phi.$

For example, a Hermitian operator or a self-adjoint operator is an operator $\phi$ such that $\phi^* = \phi$ while a unitary operator is an operator $\varphi$ such that $\varphi^* = \varphi^{-1}.$ These are examples of normal operators. Note also that a real vector space endowed with a symmetric bilinear form is a particular case of a Hermitian space. In this case, the adjoint of a matrix is the transposed matrix and symmetric matrices are Hermitian matrices while orthogonal transformations are unitary operators. As a last remark, note that if $P$ is a unitary matrix (or equivalently a change of basis matrix between two orthonormal basis) and if $M$ is normal, hermitian or unitary, then so is $P^*MP,$ so that these concepts don’t depend on the choice of basis.

From the projection formula, it is easy to see that for a space $(V, \langle \cdot , \cdot \rangle ),$ there exists an orthogonal basis $B = (v_1, \ldots, v_n)$ such that for each $i, \langle v_i, v_i \rangle$ is equal to $1, -1$ or $0.$ By Sylvester’s law (which says that the number of such 1’s, -1’s or 0’s is invariant under a change of basis), it follows that there exists an orthonormal basis for $V$ iff the form is positive definite. In this case, let $A$ be the matrix for $\langle \cdot , \cdot \rangle,$ ie $\langle u,v \rangle = u^*Av,$ let $B$ be an orthonormal basis for this form and also let $P$ be the change of basis matrix from the canonical basis $(e_1, \ldots, e_n)$ to the basis $B.$ Then we have

$\langle v_i,v_j \rangle = v_i^*Av_j = (Pe_i)^*A(Pe_j) = e_i^*(P^*AP)e_j.$

Since $\langle v_i, v_j \rangle = 1$ if $i = j$ and $0$ otherwise (because $B$ is orthonormal), we see that $P^*AP,$ the matrix for the form in the orthonormal basis, must be the identity matrix. We have proven the following proposition:

Proposition: Every positive definite Hermitian form on $V$ is conjugate to the standard Hermitian form. More precisely, if $M$ is the matrix of the form and $B$ is any orthonormal basis for $V,$ then $P^*MP = \text{Id}$ where $P$ is the change of basis matrix from the canonical basis to $B.$

Here is a characterisation that motivates the definition of normal, Hermitian and unitary operators:

Proposition: Let $\phi$ be a linear operator on a (positive definite) Hermitian space $V.$

1. For all $u,v \in V,$ we have $\langle\phi u,v \rangle = \langle u,\phi^* v \rangle$ and $\langle u, \phi v\rangle = \langle \phi^* u, v \rangle,$
2. $\phi$ is normal iff for all $u,v \in V,$ $\langle \phi u,\phi v \rangle = \langle \phi^* u, \phi^* v \rangle,$
3. $\phi$ is Hermitian iff for all $u,v \in V,$ $\langle \phi u, v \rangle = \langle v, \phi u \rangle,$
4. $\phi$ is unitary iff for all $u,v \in V,$ $\langle \phi u, \phi v \rangle = \langle u, v \rangle.$

Proof:

1. Pick an orthonormal basis $B$ for $V$ and let $M$ be the matrix of $\phi$ with respect to $B.$ If $x$ and $y$ are the respective representations of $u$ and $v$ in the basis $B,$ we have, by the last proposition,

$\langle \phi u,v \rangle = (Ax)^*y = x^*A^*y$

and

$\langle u, \phi^* v \rangle = x^* A^*y$

hence $\langle \phi u,v \rangle = \langle u, \phi^* v \rangle.$ The other formula can be proven similarly.

2.  Substituting $\phi^* u$ for $u$ in the first equation of 1., we get

$\langle \phi \phi^* u, v \rangle = \langle \phi^* u, \phi^* v \rangle.$

Substituting $\phi u$ for $u$ in the second equation, we get

$\langle \phi u, \phi v \rangle = \langle \phi^* \phi u, v \rangle.$

Therefore, if $\phi$ is normal, ie if $\phi \phi^* = \phi^* \phi,$ we have $\langle \phi u, \phi v \rangle = \langle \phi^* u, \phi^* v \rangle.$ Conversely, if $\langle \phi u, \phi v \rangle = \langle \phi^* u , \phi^* v \rangle,$ then $\langle \phi \phi^*u, v \rangle = \langle \phi^* \phi u, v \rangle.$ It follows that

$0 = \langle \phi \phi^*u, v \rangle - \langle \phi^* \phi u, v \rangle = \langle \phi \phi^*u - \phi^* \phi u, v \rangle$

Since $v$ was arbitrary and because the form is positive definite, we must have $\phi \phi^*u - \phi^* \phi u = 0,$ hence $\phi$ is normal. The proofs of 3. and 4. follow the same lines.

QED

Proposition: Let $\phi$ be a linear operator on a Hermitian space $V$ and let $W$ be a subspace of $V.$ If $W$ is $\phi$-invariant, then the orthogonal space $W^{\bot}$ is $\phi^*$-invariant. Since $(\phi^*)^* = \phi,$ if $W$ is $\phi^*$-invariant, then $W^{\bot}$ is $\phi$-invariant.

Proof: Let $w \in W$ and $u \in W^{\bot}.$ By the first point of last proposition, we have $\langle w, \phi^* u \rangle = \langle \phi w, u \rangle.$ So if $W$ is $\phi$-invariant, we have $\phi w \in W$ and thus $\langle \phi w, u \rangle = 0.$ This gives $\langle w, \phi^* u \rangle = 0$ which in other words means that $\phi^* u \in W^{\bot}$ for arbitrary $u,$ ie $W^{\bot}$ is $\phi^*$-invariant.

QED

Recall that $v$ is an eigenvector for the matrix $A$ iff $Av = \lambda v$ for some $\lambda \in \mathbf{C}$ iff $(A-\text{Id})$ is not invertible. Since $(A-\text{Id})$ is not invertible iff $\det(A-\text{Id}) = 0,$ the fundamental theorem of algebra assures us that every linear operator on a complex vector space has at least one eigenvector.

Theorem: Let $\phi$ be a normal operator on a Hermitian space $V,$ and let $v$ be an eigenvector of $\phi$ with eigenvalue $\lambda.$ Then $v$ is also an eigenvector of $\phi^*,$ with eigenvalue $\overline{\lambda}.$

Proof:

Case 1: $\lambda = 0.$ In this case, $\phi v = 0.$ Since $\phi$ is a normal operator, we have $\langle 0,0 \rangle = \langle \phi v, \phi v \rangle = \langle \phi^* v , \phi^* v \rangle.$ Since the form is positive definite, we must have $\phi^* v = 0,$ as was to be shown.

Case 2: $\lambda \neq 0.$ Then, if $\psi = \phi - \lambda \text{Id},$ $v$ is an eigenvector for $\psi$ with eigenvalue $0.$ Moreover, we have $\psi^* = \phi^* - \overline{\lambda}\text{Id}$ so

$\psi \psi^* = \phi \phi^* - \phi \overline{\lambda}\text{Id} - \phi^* \lambda \text{Id} + |\lambda|^2 \text{Id} = \psi^* \psi.$

In other words, $\psi$ is a normal operator so by case 1, $v$ is an eigenvector for $\psi^*$ with eigenvalue $0.$ ie $\phi^* v = \overline{\lambda} v,$ as wanted.

QED

In my next post, I will finish the job and prove the Spectral Theorem for Normal Operators.

Reference: M. Artin, Algebra, 2nd edition.