Throughout is an ineteger, and is an matrix with entries in

**Definition 1**. The complex conjugate of is the matrix where is the complex conjugate of Let be the transpose of

**Remark 1**. Recall that the standard inner product over the -vector space is defined by for all vectors where are the -th entry of respectively.

**Remark 2**. A direct and easy computation shows that for all Since obviously we also have for all

**Lemma**. Let and be the linear transformations corresponding to and respectively, i.e. and for all Then

*Proof*. We first remind the reader a couple of things from elementary linear algebra: if is a subspace of then where the inner product is the one defined in Remark 1. Then is a subspace of and So, to prove the lemma, we let and we will show that This is easy to do: if and only if for all So, since we have if and only if for all But, by Remark 2, we have So if and only if for all Thus if and only if i.e.

**Definition 2**. If is invertible and then is called **unitary**.

**Remark 3**. If is unitary, then for all The reason is that, by Remark 2 and the above definition, we have Using induction, we get for all integers and all

**Problem**. Suppose that is unitary and Evaluate

**Solution**. Let and be the linear transformations corresponding to and respectively and let be the identity matrix. Apply the above lemma to the matrix to get where is the identity transformation. So there exist (unique) and such that Let

Then Let’s find first. Since we have for some Thus and so, by Remark 3,

Hence

and therefore We are now going to find Since we have Thus and so Hence for all integers and so Thus and hence