# On Hermitian spaces

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****:**

A *Hermitian form* on a complex vector space is a map denoted by such that it is conjugate linear in the first argument, linear in the second, and Hermitian symmetric. Conjugate linear means that

and Hermitian symmetric means that

A Hermitian form on is said to be *positive definite* if for all with equality iff

A *Hermitian space* is a complex vector space with a positive definite Hermitian form We will only work with finite dimensional vector spaces in this post. The *standard Hermitian form* on is

Let be a linear operator on a Hermitian space and let be the matrix of with respect to an orthonormal basis The *adjoint operator* of denoted is the operator whose matrix is the adjoint of denoted (with respect to the same basis ). Recall that the adjoint of the matrix is the transposed matrix with conjugated entries. Note that this is well defined : if we change to a new orthonormal basis with a change of basis matrix the new matrix of will be Therefore its adjoint will be

which is just 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 or equivalently for the matrix of

For example, a *Hermitian operator* or a *self-adjoint operator* is an operator such that while a unitary operator is an operator such that 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 is a unitary matrix (or equivalently a change of basis matrix between two orthonormal basis) and if is normal, hermitian or unitary, then so is 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 there exists an orthogonal basis such that for each is equal to or 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 iff the form is positive definite. In this case, let be the matrix for ie let be an orthonormal basis for this form and also let be the change of basis matrix from the canonical basis to the basis Then we have

Since if and otherwise (because is orthonormal), we see that 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 is conjugate to the standard Hermitian form. More precisely, if is the matrix of the form and is any orthonormal basis for then where is the change of basis matrix from the canonical basis to

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

Proposition:Let be a linear operator on a (positive definite) Hermitian space

- For all we have and
- is normal iff for all
- is Hermitian iff for all
- is unitary iff for all

**Proof:**

**1.** Pick an orthonormal basis for and let be the matrix of with respect to If and are the respective representations of and in the basis we have, by the last proposition,

and

hence The other formula can be proven similarly.

**2. ** Substituting for in the first equation of **1.**, we get

Substituting for in the second equation, we get

Therefore, if is normal, ie if we have Conversely, if then It follows that

Since was arbitrary and because the form is positive definite, we must have hence is normal. The proofs of **3.** and **4.** follow the same lines.

QED

Proposition:Let be a linear operator on a Hermitian space and let be a subspace of If is -invariant, then the orthogonal space is -invariant. Since if is -invariant, then is -invariant.

**Proof:** Let and By the first point of last proposition, we have So if is -invariant, we have and thus This gives which in other words means that for arbitrary ie is -invariant.

QED

Recall that is an eigenvector for the matrix iff for some iff is not invertible. Since is not invertible iff the fundamental theorem of algebra assures us that every linear operator on a complex vector space has at least one eigenvector.

Theorem:Let be a normal operator on a Hermitian space and let be an eigenvector of with eigenvalue Then is also an eigenvector of with eigenvalue

**Proof:**

Case 1: In this case, Since is a normal operator, we have Since the form is positive definite, we must have as was to be shown.

Case 2: Then, if is an eigenvector for with eigenvalue Moreover, we have so

In other words, is a normal operator so by case 1, is an eigenvector for with eigenvalue ie 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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