Math 202B Lecture 5

Recalling that in Math 202B the term “vector space” means finite-dimensional complex vector space. A Hilbert space is a vector space equipped with a scalar product: a function

\langle \cdot,\cdot \rangle \colon V \times V \longrightarrow \mathbb{C}

which satisfies:

  • \langle v,\beta_1w_1+\beta_2w_2 \rangle = \beta_1\langle v,w_1\rangle + \beta_2\langle v,w_2\rangle;
  • \langle v,w \rangle = \overline{\langle w,v\rangle};
  • \langle v,v \rangle \geq 0 with equality if and only if v=0_V.

An orthonormal basis in a Hilbert space V is a vector space basis X \subset V such that \langle x,y \rangle = \delta_{xy} for all x,y \in X. The expansion of any vector v \in V with respect to an orthonormal basis is

v = \sum\limits_{x \in X} \langle x,v\rangle x,

so that the scalar product of any two vectors v,w \in V is

\langle v,w \rangle = \sum\limits_{x \in X} \overline{\langle x,v\rangle}\langle x,w\rangle.

If W is a subspace of V, its orthogonal complement is

W^\perp = \{v \in V \colon \langle v,w \rangle = 0 \text{ for all }w \in W\},

and this is a subspace of V. Furthermore,

\min(W,W^\perp) = W \cap W^\perp = \{0_V\}

and

\max(W,W^\perp) = \mathrm{Span} (W \cup W^\perp) = V.

It is assumed that the above is familiar from Math 202A.

Definition 5.1. The linear algebra of a Hilbert space V is the vector space \mathcal{L}(V) of linear operators on A \colon V \to V together with multiplication defined by composition,

[AB](v) = A(B(v)),

and conjugation defined by sending A to its adjoint A^*, which is characterized by the condition

\langle v,Aw\rangle= \langle A^*v,w\rangle, \quad v,w \in V.

The following is a deep rust removal exercise.

Problem 5.1. Prove that \mathcal{L}(V) is indeed an algebra, i.e. satisfies the conditions of Definition 1.1. Show that \mathcal{L}(V) is commutative if and only if \dim V=1.

For the remainder of this lecture we fix a Hilbert space V and an orthonormal basis X \subset V. The matrix elements of A \in \mathcal{L}(V) relative to X are the scalar products

\langle x,Ay\rangle, \quad x,y \in X.

Here is another rust removal exercise.

Problem 5.2. Prove that A=B in \mathcal{L}(V) if and only if \langle x,Ay\rangle=\langle x,By \rangle for all x,y \in X. Given A,B \in \mathcal{L}(V), compute the matrix elements of the product AB in terms of the matrix elements of its factors.

We now define a set of operators in \mathcal{L}(V) associated to the orthonormal basis X.

Definition 5.2. The elementary operators E_{xy} \in \mathcal{L}(V) are defined by

E_{xy}z = x\langle y,z\rangle, \quad x,y,z \in X.

There are |X|^2 = (\dim V)^2 elementary operators in \mathcal{L}(V). It is straightforward to calculate the matrix elements of an elementary operator: we have

\langle w,E_{xy}z\rangle = \langle w,x \langle y,z\rangle\rangle=\langle w,x\rangle \langle y,z\rangle=\delta_{wx}\delta_{yz}, \quad w,x,y,z \in X.

Proposition 5.1. The elementary operators satisfy E_{xy}^*=E_{yx} and E_{wx}E_{yz} = \langle x,y\rangle E_{wz}.

Problem 5.3. Prove Proposition 4.1.

According to Proposition 4.1, the diagonal elementary operators E_{xx}, x \in X, are orthogonal projections in \mathcal{L}(V) – they satisfy E_{xx}^* and E_{xx}E_{yy}=\delta_{xy}E_{xx}. Thus, \{E_{xx} \colon x \in X\} is a linearly independent set in \mathcal{L}(V) spanning a subalgebra \mathcal{D}(X) of \mathcal{L}(V) isomorphic to \mathcal{F}(X) via E_{xx} \mapsto E_x. We call \mathcal{D}(X) the diagonal subalgebra of \mathcal{L}(V) relative to the orthonormal basis X \subset V. If A \in \mathcal{D}(X), we say that A acts diagonally on the orthonormal basis X, because the the fact that

A=\sum\limits_{x \in X} \alpha_x E_{xx}

means that only the diagonal matrix elements of A can be nonzero. Equivalently, \mathcal{D}(X) is the subalgebra of \mathcal{L}(V) consisting of all operators whose matrix relative to X is diagonal, and the diagonal matrix elements of every such A define a function on X.

Proposition 5.2. The set \{E_{xy} \colon x,y \in X\} of elementary operators spans \mathcal{L}(V).

Proof: Let A \in \mathcal{L}(V) be any operator, and let a_{xy} = \langle x,Ay\rangle be its matrix elements relative to the orthonormal basis X. Define B \in \mathcal{L}(V) by

B=\sum\limits_{x,y \in X} a_{xy}E_{xy}.

We claim that A=B. Indeed, the matrix elements of B are

\langle w,Bz\rangle = \sum\limits_{x,y \in X} a_{xy}\langle w,E_{xy}z\rangle = a_{wz}, \quad w,z \in X.

Thus, A=B by Problem 4.1.

-QED

We will now prove that \{E_{xy} \colon x,y \in X\} is a linearly independent set in \mathcal{L}(V), hence a linear basis. We will do this by introducing a scalar product on \mathcal{L}(V) with respect to which \{E_{xy} \colon x,y \in X\} is an orthonormal set.

Definition 5.3. The trace on \mathcal{L}(V) relative to X is the linear functional defined by

\mathrm{Tr}_X(A) = \sum\limits_{x \in X} \langle x,Ax\rangle, \quad A \in \mathcal{L}(V).

Proposition 5.3. We have

  • \mathrm{Tr}_X(I) = \dim V;
  • \mathrm{Tr}_X(AB)=\mathrm{Tr}_X(BA);
  • \mathrm{Tr}_X(A^*A) \geq 0 with equality if and only if A=0.

Proof: First, we have

\mathrm{Tr}_X(I) = \sum\limits_{x \in X} \langle x,Ix\rangle = \sum\limits_{x\in X}1 = |X|=\dim V.

Second, we have

\mathrm{Tr}_X(AB)=\sum\limits_{x\in X} \langle x,ABx\rangle=\sum\limits_{x \in X} \langle x,A\sum\limits_{y \in X} \langle y,Bx\rangle y\rangle=\sum\limits_{x \in X}\sum\limits_{y\in X} \langle x,Ay\rangle \langle y,Bx \rangle,

and changing order of summation this is

\sum\limits_{y\in X}\sum\limits_{x \in X}\langle y,Bx \rangle\langle x,Ay\rangle=\sum\limits_{y \in X} \langle y,B\sum\limits_{x \in X} \langle x,Ay\rangle x\rangle=\sum\limits_{y \in X} \langle y,BAy\rangle=\mathrm{Tr}_X(BA).

Third, we have

\mathrm{Tr}_X(A^*A) = \sum\limits_{x \in X} \langle x,A^*Ax\rangle = \sum\limits_{x \in X} \langle Ax,Ax\rangle = \sum\limits_{x\in X} \|Ax\|^2,

which vanishes if and only if every term of the sum vanishes, and this occurs if and only if A maps every vector in the basis X to the zero vector.

-QED

Proposition 5.4. Let X and Y be two orthonormal bases of V. We have \mathrm{Tr}_X=\mathrm{Tr}_Y.

Proof: For any enumeration x_1,\dots,x_n of X, and any enumeration y_1,\dots,y_n of Y, the linear operator U \colon V \to V defined by Ux_i=y_i is unitary. We have

\mathrm{Tr}_Y(A) = \sum\limits_{i=1}^n \langle y_i,Ay_i\rangle=\sum\limits_{I=1}^n \langle Ux_i,AUx_i\rangle = \mathrm{Tr}_X(U^*AU).

By Proposition 14.2, we have \mathrm{Tr}_X(U^*AU)=\mathrm{Tr}_X(AUU^*)=\mathrm{Tr}_X(A).

-QED

Proposition 14.3 shows that Definition 14.3 gives the same linear functional on \mathcal{L}(V) no matter what orthonormal basis X is used to define it, and we call this functional the trace on V.

Problem 5.4. Show that \mathrm{Tr} E_{xy}=\langle x,y \rangle.

Recall the following basic notion from Math 202A.

Definition 5.4. The spectrum of A \in \mathcal{L}(V) is the set \mathrm{Spec}(A) of numbers \alpha \in \mathbb{C} such that \alpha I-A is not invertible. Elements of \mathrm{Spec}(A) are called eigenvalues of A.

Because \alpha I - A being non-invertible is equivalent to \det(\alpha I - A)=0, and because \det (\alpha I-A) is a polynomial in \alpha, the Fundamental Theorem of Algebra says that \mathrm{Spec}(A) is nonempty for every A \in \mathcal{L}(V).

More recall from Math 202A: a number \alpha\in is an eigenvalue of an operator A \in \mathcal{L}(V) if and only if the kernel of \alpha I-A is not the zero subspace of \mathcal{L}(V). The nonzero subspace \mathrm{Ker}(\alpha I-A) is called the \alpha–eigenspace of A, and any nonzero vector v \in \mathrm{Ker}(\alpha I-A) is said to be an eigenvector of A corresponding to the eigenvalue \alpha.

Problem 5.5. Prove that for any A \in \mathcal{L}(V) we have

\mathrm{Tr}\, A = \sum\limits_{a \in \mathrm{Spec}(A)} a.

Now we use the trace to define a scalar product on \mathcal{L}(V).

Definition 5.5. The Frobenius scalar product on \mathcal{L}(V) is defined by \langle A,B \rangle = \mathrm{Tr} A^*B.

The definition of the Frobenius scalar product should be familiar from Math 202A. The associated norm is called the Frobenius norm, \|A\|=\sqrt{\mathrm{Tr A^*A}}.

Problem 5.6. Show that the Frobenius scalar product is indeed a scalar product, and moreover that it interacts with multiplication and conjugation in \mathcal{L}(V) according to \langle AB,C\rangle =\langle B,A^*C\rangle=\langle A,CB^*\rangle for all A,B,C \in \mathcal{L}(V).

Theorem 5.5. The elementary operators \{E_{xy}\colon (x,y) \in X \times X\} form an orthonormal basis of \mathcal{L}(V) relative to the Frobenius scalar product.

Proof: We have already shown that \{E_{xy}\colon (x,y) \in X \times X\} spans \mathcal{L}(V), so it remains to show this is an orthonormal set with respect to the Frobenius scalar product. For any w,x,y,z \in X we have

\langle E_{wx},E_{yz}\rangle = \mathrm{Tr}(E_{wx}^*E_{yz})=\mathrm{Tr}(E_{xw}E_{yz}) = \langle w,y\rangle\mathrm{Tr}(E_{xz})=\langle w,y\rangle \langle x,z\rangle.

-QED

Problem 5.7. Prove that the diagonal subalgebra \mathcal{D}(X) is a maximal abelian subalgebra of \mathcal{L}(V).

Two Hilbert spaces V and W are said to be isomorphic if there is a linear bijection between them.

The trace on \mathcal{L}(V) is unique in the following sense.

Theorem 5.6. If \tau \colon \mathcal{L}(V) \to \mathbb{C} is a linear functional which satisfies \tau(AB)=\tau(BA) for all A,B \in \mathcal{L}(V), then \tau is a scalar multiple of the trace.

Proof: Since \tau is linear, it is uniquely determined by its values on the elementary basis E_{xy} of \mathcal{L}(V).

For distinct x,y \in X, we have

\tau(E_{xy})=\tau(E_{xy}E_{yy}) = \tau(E_{yy}E_{xy})=\tau(0)=0,

so \tau is zero on the off-diagonal elementary operators. For the diagonal elementary operators,

\tau(E_{xx})=\tau(E_{xy}E_{yx})=\tau(E_{yx}E_{xy})=\tau(E_{yy}).

Thus, \tau is constant on the diagonal elementary operators, i.e. there is a scalar c such that \tau(E_{xx})=c for all x \in X. Thus for any

A = \sum\limits_{x,y \in X} \alpha_{xy} E_{xy},

we have

\tau(A) = \sum\limits_{x \in X} \tau(E_{xx}) = c\dim V = c\mathrm{Tr} A.

-QED

Problem 5.8. Let V be a Hilbert space of dimension at least two. Show that there does not exist an algebra homomorphism \mathcal{L}(V) \to \mathbb{C}.

Published by Jonathan Novak

Mathematician at UC San Diego.

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