# 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.