Equivalent Conditions to be a Unitary Matrix

Problem 29

A complex matrix is called unitary if $\overline{A}^{\trans} A=I$.

The inner product $(\mathbf{x}, \mathbf{y})$ of complex vector $\mathbf{x}$, $\mathbf{y}$ is defined by $(\mathbf{x}, \mathbf{y}):=\overline{\mathbf{x}}^{\trans} \mathbf{y}$. The length of a complex vector $\mathbf{x}$ is defined to be $||\mathbf{x}||:=\sqrt{(\mathbf{x}, \mathbf{x})}$.

Let $A$ be an $n \times n$ complex matrix. Prove that the followings are equivalent.

(a) The matrix $A$ is unitary.

(b) $||A \mathbf{x}||=|| \mathbf{x}||$ for any $n$-dimensional complex vector $\mathbf{x}$.

(c) $(A\mathbf{x}, A\mathbf{y})=(\mathbf{x}, \mathbf{y})$ for any $n$-dimensional complex vectors $x, y$

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Steps.

Try to show the implications (a)$\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a).

Proof.

We will show that (a)$\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a).


(a) $\Rightarrow$ (b).

Suppose that $A$ is unitary. Then we have for any $n$-dimensional complex vector $\mathbf{x}$
\begin{align*}
||A\mathbf{x}|| &=\sqrt{ {(\overline{A\mathbf{x}})^{\trans}} (A \mathbf{x})}
=\sqrt{ \overline{\mathbf{x}}^{\trans} \overline{A}^{\trans} A\mathbf{x} } \\
&= \sqrt{ \overline{\mathbf{x}}^{\trans} \mathbf{x} }=||\mathbf{x}||.
\end{align*}

(b) $\Rightarrow$ (c).

Suppose that the statement (b) is true.
Then we compute $||A(\mathbf{x}+\mathbf{y})||^2=||\mathbf{x}+\mathbf{y}||^2$.
The left hand side is
\begin{align*}
(A\mathbf{x}+A\mathbf{y} ,A\mathbf{x}+A\mathbf{y})&=(A\mathbf{x},A\mathbf{x})+(A\mathbf{x},A\mathbf{y})+(A\mathbf{y},A\mathbf{x})+(A\mathbf{y}, A\mathbf{y}) \\
&=||A\mathbf{x}||^2 + (A\mathbf{x},A\mathbf{y}) +\overline{(A\mathbf{x},A\mathbf{y})} + ||A\mathbf{y}||.
\end{align*}
The right hand side is
\begin{align*}
(\mathbf{x}, \mathbf{x})+(\mathbf{x}, \mathbf{y})+(\mathbf{y}, \mathbf{x})+(\mathbf{y}, \mathbf{y})=||\mathbf{x}||+(\mathbf{x}, \mathbf{y}) +\overline{(\mathbf{x}, \mathbf{y})} +||\mathbf{y}||.
\end{align*}
Using (b) we cancel the length terms and we obtain the relation
\begin{align*}
(A\mathbf{x},A\mathbf{y}) +\overline{(A\mathbf{x},A\mathbf{y})}=(\mathbf{x}, \mathbf{y}) +\overline{(\mathbf{x}, \mathbf{y})}. \tag{*}
\end{align*}
that holds for any $\mathbf{x}$ and $\mathbf{y}$.

This implies that we have the equality of the real parts $\Repart (A\mathbf{x},A\mathbf{y})= \Repart (\mathbf{x}, \mathbf{y})$.
To show that the imaginary parts are also equal, we substitute $ix$ into $x$ in the equation (*). (Here $i=\sqrt{-1}$.)
Then we have
\begin{align*}
i\left((A\mathbf{x},A\mathbf{y}) -\overline{(A\mathbf{x},A\mathbf{y})} \right)=i\left( (\mathbf{x}, \mathbf{y}) -\overline{(\mathbf{x}, \mathbf{y})} \right). \end{align*}
Thus, we have $\Impart( A\mathbf{x},A\mathbf{y})=\Impart (\mathbf{x}, \mathbf{y})$.
Hence we have $(A\mathbf{x},A\mathbf{y})=(\mathbf{x}, \mathbf{y})$.

(c) $\Rightarrow$ (a).

We give two proofs of this implication.

By assumption (c), we have $(A\mathbf{x},A\mathbf{y})=(\mathbf{x}, \mathbf{y})$ for any $\mathbf{x}$ and $\mathbf{y}$. Equivalently we have
\begin{align*}
\overline{\mathbf{x}}^{\trans}\overline{A}^{\trans}A \mathbf{y}=(\mathbf{x}, \mathbf{y}) \tag{**}.
\end{align*}
Let $e_i$ be a unit $n$-dimensional vector whose entries are all zero except that the $i$-th entry is $1$.
Take $\mathbf{x}=e_i$ and $\mathbf{y}=e_j$, we see that the left hand side of (**) is the $(i,j)$-entry of the matrix $\overline{A}^{\trans}A$ and the right hand side of (**) is $\delta_{i,j}$. Thus we have $\overline{A}^{\trans}A=I_n$.

The 2nd Proof.

We have for any $\mathbf{x}$ and $\mathbf{y}$,
\begin{align*}
(\mathbf{x}, (\overline{A}^{\trans} A -I_n) \mathbf{y}) &= (\mathbf{x}, \overline{A}^{\trans}A \mathbf{y})-(\mathbf{x}, \mathbf{y}) \\
&=(A \mathbf{x}, A\mathbf{y})-(\mathbf{x}, \mathbf{y})=0.
\end{align*}
Therefore, we have $\overline{A}^{\trans} A =I_n$.

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