## A Fun Interpretation of Unitary Matrices2022-05-28 ☾
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 \(x_i^* x_j = \delta_{ij}\). (This is the Kronecker delta. It equals \(1\) if \(x_i = x_j\) and \(0\) otherwise.) 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: \[ (U^*U)_{ij} = x_i^* x_j = \delta_{ij} \Longrightarrow U^*U = I \Longrightarrow UU^* = I \,, \] where the last identity follows from the fact that if \(AB=I\), then \(BA=I\). (See the proof here. Note that the class of square matrices \(U\) adhering to the identity \( U^*U = UU^* = I \) are the unitary matrices.)
But this is interesting… the last identity implies that the To answer this, it helps think about matrix multiplication as a series of \[ UU^* = \sum_{i=1}^n x_i x_i^* = P_{x_1} + \cdots + P_{x_n} = I \,. \] 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. |