In Lecture 14, we considered a special type of linear operators known as orthogonal transformation, which were defined as follows. Let be an -dimensional Euclidean space. An operator is orthogonal if the image of any orthonormal basis of is again an orthonormal basis of We found that orthogonal operators can alternatively be characterized as those linear operators which preserve the scalar product, meaning that

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

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

**Theorem 1:** For any operator there is a unique operator such that

*Proof:* We first prove that an operator with the desired property exists. Let be an orthonormal basis in Let be the operator defined by

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

which is equivalent to saying that the -element of the matrix is equal to the -element of the matrix a relationship which is usually expressed as saying that is the transpose of Now, for any vectors we have

Now we prove uniqueness. Suppose that are two operators such that

Then, we have that

In particular, we have

or in other words that is the zero matrix. Since the map which takes an operator to its matrix relative to is an isomorphism, this means that is the zero operator, or equivalently that

— Q.E.D.

**Definition 1:** For each we denote by the unique operator such that for all . We call the **adjoint** of

According to Definition 1, yet another way to express that is an orthogonal operator is to say that the adjoint of is the inverse of i.e. 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 is said to be **selfadjoint** if .

Owing to the fact that a selfadjoint operator satisfies

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 is said to be **normal** if it commutes with its adjoint, i.e.

**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 is an orthogonal operator, then and also If is a selfadjoint operator, then and also

— Q.E.D.

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