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On Hermitian spaces

May 14, 2012

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.

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