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 $$x_i^* x_j = \delta_{ij}$$. 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 $$BA=I$$. 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?

  This is the Kronecker delta. It equals $$1$$ if $$x_i = x_j$$ and $$0$$ otherwise.
  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.