Exercise 11
Let \(A\) be an \(n \times n\) matrix with complex entries. Prove that \(AA^* = I\) if and only if the rows of \(A\) form an orthonormal basis for \(\mathbb{C}^n\).
Proof:
By definition we know that \(A^*\) is defined as \((A^*)_{ij} = \overline{A_{ji}}\) for all \(i\) and \(j\). Therefore the row \(i\), column \(j\) entry of \(AA^*\) is
$$
\begin{align*}
(AA^*)_{ij} &= \sum_{k=1}^n A_{ik}(A^*)_{kj} \\
&= \sum_{k=1}^n A_{ik}\overline{A_{jk}} \\
&= A_{i1}\overline{A_{j1}} + A_{i2}\overline{A_{j2}} + ... + A_{in}\overline{A_{jn}} \\
&= \langle A_i, A_j \rangle
\end{align*}
$$
where \(A_i\) and \(A_j\) are the \(i\)th and \(j\)th rows of \(A\). What does having \(AA^{*} = I\) then mean? The \(i\)th row and \(j\)th column entry of \(I\) is \(I_{ij} = \delta_{ij}\) and this corresponds to the inner product of the rows \(A_i\) and \(A_j\).
$$
\begin{align*}
\begin{pmatrix}
\cdots & \cdots & \cdots \\
\cdots & A_{i} & \cdots \\
\cdots & \cdots & \cdots \\
\end{pmatrix}
\begin{pmatrix}
\cdots & \cdots & \cdots \\
\cdots & A_{j} & \cdots \\
\cdots & \cdots & \cdots \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & \cdots & 0 \\
0 & 1 = \delta_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{pmatrix}
\end{align*}
$$
So \(AA^{*} = I\) is equivalent to having \(\langle A_i, A_j \rangle = \delta_{ij}\). But this implies that the rows of \(A\) form an orthonormal set of \(n\) vectors in \(C^{n}\) by definition of an orthonormal set. Since there are \(n\) of them and they are linearly independent, then they form a basis for \(C^{n}\). \(\ \blacksquare\).
References