Complex square matrices, proving there exists x,y in C^n such that A=xy*

Question

Suppose A $\in$ $M_{nxn}$ the set of complex square matrices. Show the following statements are equivalent

a) A has rank 1

b) $\exists$x,y $\in$ $C^n$ such that $A=xy^*$

What are the right and left eigenvectors of A and the corresponding eigenvalues?

Answer 1

Hopefully you find $b \implies a$ clear, so $a \implies b$ is what we need to prove

There are a few approaches to this problem.

Option 1: use the SVD.

Let $A$ be a rank-1 square matrix of size $n$. We can write $$ A = U\Sigma V^* $$ Let $u_1,\dots,u_n$ denote the columns of $U$ and $v_1,\dots, v_n$ the columns of $V$. We can write $$ A = \pmatrix{u_1 & \cdots & u_n} \pmatrix{\sigma_1\\&0\\&&\ddots\\&&&0} \pmatrix{v_1^* \\ \vdots \\ v_n^*} $$ computing this product by block-matrix multiplication yields $$ A = \sigma_1 u_1 v_1^* $$

In fact, this is another way of looking at SVD. In general, a rank $k$ matrix can be written as $$ A = \sum_{i=1}^k \sigma_i u_i v_i^* $$

Option 2: See Jimmy K's hint in his comment