Math 202B: Lecture 18

In this Lecture, we develop some further purely linear-algebraic background which we will use to study unitary representations of finite groups. Most of this will probably be familiar from Math 202A albeit stated in a possibly different way.

Let V and W be Hilbert spaces of positive, finite dimension. We denote the scalar product in V by \langle \cdot,\cdot \rangle_V and the scalar product in W by \langle \cdot,\cdot \rangle_W. Furthermore, let Y \subset V be an orthonormal basis of V, and let X \subset W be an orthonormal basis of W.

Proposition 18.1. For each T \in \mathrm{Hom}(V,W), there exists a unique T^* \in \mathrm{Hom}(W,V), called the adjoint of T, such that

\langle w,Tv \rangle_W = \langle T^*w,v \rangle_V, \quad v\in V,\ w \in W.

Problem 18.1. Prove Proposition 18.1.

The adjoint operation leads to a natural scalar product on the vector space \mathrm{Hom}(V,W).

Definition 18.2. The Frobenius scalar product on \mathrm{Hom}(V,W) is defined by

\langle S,T \rangle = \mathrm{Tr}\, S^*T, \quad S,T \in \mathrm{Hom}(V,W).

Note that the Frobenius scalar product \langle \cdot,\cdot \rangle carries no subscript, differentiating it from the scalar products \langle \cdot,\cdot \rangle_V and \langle \cdot,\cdot\rangle_W in V and W.

Proposition 18.2. The Frobenius scalar product is indeed a scalar product on \mathrm{Hom}(V,W).

Proof: We have

\langle S,T \rangle = \sum\limits_{y \in Y} \langle y,S^*Ty\rangle_V = \sum\limits_{y \in Y} \langle Sy,Ty \rangle_W.

Linearity in the second slot is easily verified:

\langle S,\beta_1T_1 +\beta_2T_2\rangle = \beta_1\langle S,T_1\rangle+\beta_2\langle S,T_2\rangle,

as is \langle T,S \rangle = \overline{\langle S,T\rangle}. As for positive definiteness, we have

\langle T,T \rangle = \sum\limits_{y \in Y} \langle Ty,Ty\rangle_W = \sum\limits_{y \in Y} \|Ty\|_W^2,

and this sum of nonnegative numbers is zero if and only if all its terms are zero, which happens if and only if Ty=0_W for all y \in Y, which is the case if and only if T \in \mathrm{Hom}(V,W) is the zero transformation.

-QED

Now that \mathrm{Hom}(V,W) has been promoted from a vector space to a Hilbert space, we seek an orthonormal basis in this space. Let Y \subset V and X\subset W be orthonormal bases of the underlying spaces which define \mathrm{Hom}(V,W).

Definition 18.3. The elementary transformations in \mathrm{Hom}(V,W) relative to X,Y are defined by

E_{xy}y' =x\langle y,y'\rangle_V, \quad x \in X,\ y,y'\in Y.

Note that if V=W and X=Y, then the elementary transformations E_{xy} \in \mathrm{Hom}(V,V)=\mathcal{L}(V) are the elementary operators in \mathcal{L}(V) discussed previously. However if V and W are different Hilbert spaces, the transformations E_{xy} \in \mathrm{Hom}(V,W) and E_{yx} \in \mathrm{Hom}(W,V) belong to different vector spaces.

Proposition 18.3. The elementary transformations satisfy E_{xy}^* = E_{yx}.

Proof: Let (x,y) \in X \times Y \subset W \times V, and let E_{xy} \in \mathrm{Hom}(V,W) be the corresponding elementary transformation. For any (x',y') \in X \times Y, we have

\langle E_{xy}y',x'\rangle_W = \langle x\langle y,y'\rangle_V,x'\rangle_W = \langle x,x'\rangle_W \langle y,y'\rangle_V

and

\langle y',E_{yx}x'\rangle_V = \langle y',y\langle x,x'\rangle_W\rangle_V=\langle x,x'\rangle_W \langle y',y\rangle_V.

-QED

Proposition 18.4. The set \{E_{xy} \colon (x,y) \in X \times Y\} is an orthonormal basis of \mathrm{Hom}(V,W) with respect to the Frobenius scalar product.

Proof: For any x,x' \in X and y,y' \in Y, we have

\langle E_{xy},E_{x'y'}\rangle = \sum\limits_{z \in Y} \langle E_{xy}z,E_{x'y'}z\rangle = \sum\limits_{z \in Y}\langle x\langle y,z\rangle,x'\langle y',z\rangle\rangle= \langle x,x'\rangle\sum\limits_{z \in Y} \langle y,z\rangle \langle y',z\rangle =\langle x,x'\rangle \langle y,y'\rangle,

which establishes orthonormality of the elementary transformations in \mathrm{Hom}(V,W) with respect to the Frobenius scalar product. Verification of the fact that \{E_{xy} \colon (x,y) \in X \times Y\} spans \mathrm{Hom}(V,W) is left as an exercise.

-QED

Problem 18.2. Prove that for any A \in \mathrm{Hom}(V,W) and B \in \mathrm{Hom}(W,V) we have \mathrm{Tr}\, AB=\mathrm{Tr}\, BA. Note that this is more surprising than in the case of square matrices, since the operators AB \in \mathcal{L}(W) and BA \in \mathcal{L}(V) act in different spaces.

Now that \mathrm{Hom}(V,W) is a Hilbert space with an explicit orthonormal basis, we are in position to consider the algebra \mathcal{L}(\mathrm{Hom}(V,W)) of linear operators on this Hilbert space (i.e. linear operators on linear transformations), and to concretely calculate the matrices of such operators. In particular, we consider the map

\omega \colon \mathcal{L}(V) \times \mathcal{L}(W) \longrightarrow \mathcal{L}(\mathrm{Hom}(V,W))

defined by

\omega(A,B)T = BTA^*, \quad T \in \mathrm{Hom}(V,W).

First of all, \omega(A,B)T is a composition of linear transformations V \to V \to W \to W, so it is indeed a linear transformation V \to W, and the map \omega is itself sesquilinear.

Proposition 18.45 If A \in \mathcal{L}(V) and B \in \mathcal{L}(W) are unitary, then \omega(A,B) \in \mathcal{L}(\mathrm{Hom}(V,W)) is unitary.

Proof: For any S,T \in \mathrm{Hom}(V,W), we have

\langle \omega(A,B)S,\omega(A,B)T\rangle = \langle BSA^*,BTA^*\rangle=\mathrm{Tr}\, (BSA^*)*(BTA^*)=\mathrm{Tr}\, AS^*B^*BTA^*=\mathrm{Tr}\, AS^*TA^*=\mathrm{Tr}\,S^*TA^*A= \langle A,B\rangle.

-QED

You have in fact seen this construction before, in Math 202A, in the context of the Singular Value Decomposition of a linear transformation.

Proposition 18.6. For any A \in \mathcal{L}(V) and B \in \mathcal{L}(W), the matrix elements of the operator \omega(A,B) \in \mathcal{L}(\mathrm{Hom}(V,W)) relative to the orthonormal basis \{E_{xy} \colon X \times Y\} are

\langle E_{xy},\omega(A,B)E_{x'y'}\rangle = \overline{\langle y,Ay'\rangle}\langle x,Bx'\rangle.

Proof: We have

\langle E_{xy},\omega(A,B)E_{x'y'}\rangle = \mathrm{Tr} E_{xy}^*BE_{xy}A^*=\mathrm{Tr}BE_{x'y'}A^*E_{xy}^*.

The trace on the right hand side is

\mathrm{Tr}BE_{x'y'}A^*E_{xy}^*=\sum\limits_{w \in X} \langle w,BE_{x'y'}A^*E_{xy}^*w\rangle=\sum\limits_{w \in X} \langle w,BE_{x'y'}A^*y\rangle\langle x,w\rangle=\langle x,BE_{x'y'}A^*y\rangle.

The scalar product on the right is

\langle x,BE_{x'y'}A^*y\rangle=\langle x,Bx'\rangle \langle y',A^*y\rangle = \langle x,Bx'\rangle \langle Ay',y\rangle=\overline{\langle y,Ay'\rangle} \langle x,Bx'\rangle.

-QED

Published by Jonathan Novak

Mathematician at UC San Diego.

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