2022-05-28 | *Why are the rows orthonormal?*

This is a proof that if the columns of the matrix are orthonormal, then the rows are also orthonormal.

Let \(U \in M_n(\mathbb{C})\) be a square matrix such that its columns form an orthonormal basis over \(\mathbb{C}^n\). That is, for each pair of columns \(x_i\) and \(x_j\), the inner product \(x_i^* x_j = \delta_{ij}\).^{[1]} If we think about matrix multiplication as a series of inner products between the rows of the first matrix and the columns of the second matrix, then this gives us the following identities:

where the last identity follows from the fact that if \(AB=I\), then \(BA=I\).^{[2]} But this is interesting... the last identity implies that the *rows* of \(U\) are also orthonormal! **If the columns of the matrix are orthonormal, then why are the rows also orthonormal?**

[1] | This is the Kronecker delta. It equals \(1\) if \(x_i = x_j\) and \(0\) otherwise. |

[2] | See the proof here. Note that the class of square matrices \(U\) adhering to the identity \( U^*U = UU^* = I \,, \) are the unitary matrices. |

Here's why the last equality holds: since the vectors \(\{x_1, \dots, x_n\}\) form a basis over \(\mathbb{C}^n\), the sum of \(n\) rank-one projection matrices forms a projection into \(\mathbb C^n \,.\) Therefore, multiplying a vector \(y \in \mathbb C^n\) by the matrix \(UU^*= (P_{x_1} + \cdots + P_{x_n})\) results in no loss of dimensionality since we are projecting onto the full vector space \(\mathbb C^n\). That is,

\[ \begin{align} UU^* y &= (P_{x_1} + \cdots + P_{x_n}) y \\ &= P_{x_1} y + \cdots + P_{x_n} y \\ &= c_1 x_1 + \cdots + c_n x_n \\ &= y \,, &&\text{(It's just a change of basis!)} \end{align} \]implying that

\[UU^* = I \,.\]This proof shows that orthonormal columns imply orthonormal rows, giving another intuition for the geometric behavior of unitary matrices.

© Vivek Gopakrishnan. Last modified: September 16, 2022. Website built with Franklin.jl and the Julia programming language.