Math 31AH: Lecture 16

In Lecture 14, we considered a special type of linear operators known as orthogonal transformation, which were defined as follows. Let \mathbf{V} be an n-dimensional Euclidean space. An operator U \in \mathrm{End}\mathbf{V} is orthogonal if the image \{U\mathbf{e}_1,\dots,U\mathbf{e}_N\} of any orthonormal basis \mathbf{e}_1,\dots,\mathbf{e}_N\} of \mathbf{V} is again an orthonormal basis of \mathbf{V}. We found that orthogonal operators can alternatively be characterized as those linear operators which preserve the scalar product, meaning that

\langle U\mathbf{v},U\mathbf{w} \rangle = \langle \mathbf{v},\mathbf{w} \rangle \quad\forall \mathbf{v},\mathbf{w} \in \mathbf{W}.

Yet another way to characterize orthogonal operators is to say that they are invertible, and

\langle \mathbf{v},U\mathbf{w} \rangle = \langle U^{-1}\mathbf{v},\mathbf{w}\rangle \quad\forall \mathbf{v},\mathbf{w} \in \mathbf{W}.

This last characterization makes contact with a more general operation on operators.

Theorem 1: For any operator A \in \mathrm{End}\mathbf{V}, there is a unique operator B \in \mathrm{End}\mathbf{V} such that

\langle \mathbf{v},A\mathbf{w} \rangle = \langle B\mathbf{v},\mathbf{w}\rangle \quad\forall \mathbf{v},\mathbf{w} \in \mathbf{W}.

Proof: We first prove that an operator with the desired property exists. Let E=\{\mathbf{e}_1,\dots,\mathbf{e}_n\} be an orthonormal basis in \mathbf{V}. Let B \in \mathrm{End}\mathbf{V} be the operator defined by

B\mathbf{e}_j = \sum_{i=1}^n \langle A\mathbf{e}_i,\mathbf{e}_j \rangle \mathbf{e}_i, \quad 1 \leq j \leq n.

That is, we have defined the operator B in such a way that its matrix elements relative to the basis E satisfy

\langle \mathbf{e}_i,B\mathbf{e}_j \rangle = \langle A\mathbf{e}_i,\mathbf{e}_j \rangle = \langle \mathbf{e}_j,A\mathbf{e}_i\rangle,

which is equivalent to saying that the (i,j)-element of the matrix [B]_E is equal to the (j,i)-element of the matrix [A]_E, a relationship which is usually expressed as saying that [B]_E is the transpose of [A]_E. Now, for any vectors \mathbf{v},\mathbf{w} \in \mathbf{V}, we have

\langle \mathbf{v},A\mathbf{w} \rangle \\ = \left \langle \sum_{i=1}^n \langle \mathbf{e}_i,\mathbf{v} \rangle \mathbf{e}_i,A\sum_{j=1}^n \langle \mathbf{e}_j,\mathbf{w} \rangle \mathbf{e}_j \right\rangle \\ = \left \langle \sum_{i=1}^n \langle \mathbf{e}_i,\mathbf{v} \rangle \mathbf{e}_i,\sum_{j=1}^n \langle \mathbf{e}_j,\mathbf{w} \rangle A\mathbf{e}_j \right\rangle \\ = \sum_{i,j=1}^n \langle \mathbf{e}_i,\mathbf{v} \rangle \langle \mathbf{e}_j,\mathbf{w}\rangle \langle \mathbf{e}_i,A\mathbf{e}_j \rangle \\ = \sum_{i,j=1}^n \langle \mathbf{e}_i,\mathbf{v} \rangle \langle \mathbf{e}_j,\mathbf{w}\rangle \langle B\mathbf{e}_i,\mathbf{e}_j \rangle \\ = \left \langle \sum_{i=1}^n \langle \mathbf{e}_i,\mathbf{v} \rangle B\mathbf{e}_i,\sum_{j=1}^n \langle \mathbf{e}_j,\mathbf{w} \rangle \mathbf{e}_j \right\rangle \\ = \left \langle B\sum_{i=1}^n \langle \mathbf{e}_i,\mathbf{v} \rangle \mathbf{e}_i,\sum_{j=1}^n \langle \mathbf{e}_j,\mathbf{w} \rangle \mathbf{e}_j \right\rangle \\ = \langle B\mathbf{v},\mathbf{w}\rangle.

Now we prove uniqueness. Suppose that B,C \in \mathrm{End}\mathbf{V} are two operators such that

\langle \mathbf{v},A\mathbf{w} \rangle = \langle B\mathbf{v},\mathbf{w} \rangle = \langle C\mathbf{v},\mathbf{w} \rangle \quad \forall \mathbf{v},\mathbf{w} \in \mathbf{V}.

Then, we have that

\left\langle (B-C)\mathbf{v},\mathbf{w} \right\rangle = 0 \quad\forall\mathbf{v},\mathbf{w} \in \mathbf{V}.

In particular, we have

\left\langle (B-C)\mathbf{e}_j,\mathbf{e}_i \right\rangle = 0 \quad 1 \leq i,j \leq n,

or in other words that [B-C]_E is the zero matrix. Since the map which takes an operator to its matrix relative to E is an isomorphism, this means that B-C is the zero operator, or equivalently that B=C.

— Q.E.D.

Definition 1: For each A \in \mathrm{End}\mathbf{V}, we denote by A^* the unique operator such that \langle A^*\mathbf{v},\mathbf{w} \rangle = \langle \mathbf{v},A\mathbf{w} \rangle for all \mathbf{v},\mathbf{w} \in \mathbf{V}. We call A^* the adjoint of A.

According to Definition 1, yet another way to express that U \in \mathrm{End}\mathbf{V} is an orthogonal operator is to say that the adjoint of U is the inverse of U, i.e. U^*=U^{-1}. Operators on Euclidean space which are related to their adjoint in some predictable way play a very important role in linear algebra.

Definition 2: An operator X \in \mathrm{End}\mathbf{V} is said to be selfadjoint if X^*=X.

Owing to the fact that a selfadjoint operator X satisfies

\langle\mathbf{v},X\mathbf{w} \rangle = \langle X\mathbf{v},\mathbf{w} \rangle \quad \forall \mathbf{v},\mathbf{w} \in \mathbf{V},

selfadjoint operators are also often called symmetric operators. The matrix of a selfajdoint operator in any basis is equal to its own transpose. We are going to study selfadjoint operators extensively in the coming lectures.

Defintion 3: An operator A \in \mathrm{End}\mathbf{V} is said to be normal if it commutes with its adjoint, i.e. A^*A=AA^*.

Proposition 1: Orthogonal operators are normal operators, and selfadjoint operators are normal operators.

Proof: This is very straightforward, but worth going through at least once. If U is an orthogonal operator, then U^*U= U^{-1}U=I, and also UU^* = UU^{-1}=I. If X is a selfadjoint operator, then X^*X=XX=X^2, and also XX^*=XX=X^2.

— Q.E.D.

Lecture 16 coda

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