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 \(\mathbf{C}^n\).


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