Vivek Gopalakrishnan

A Fun Interpretation of Unitary Matrices

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:

\[(U^*U)_{ij} = x_i^* x_j = \delta_{ij} \Longrightarrow U^*U = I \Longrightarrow UU^* = I \,,\]

where the last identity follows from the fact that if \(AB=I\), then \(BA=I\).[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.
To answer this, it helps think about matrix multiplication as a series of outer products: \( U^*U = \sum_{i=1}^n x_i x_i^* \,. \) For a pair of orthonormal vectors \(x_i\) and \(x_j\), the rank-one matrix \( P_{x_i} = x_i x_i^* \) represents the orthogonal projection onto the vector \(x_i\). Therefore,

\[ UU^* = \sum_{i=1}^n x_i x_i^* = P_{x_1} + \cdots + P_{x_n} = I \,. \]

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.