Let $A$ be an $m\times n$ matrix with $\text{rank}(A) = 1$. Prove that for some column $u$ in $\mathbb{R}^m$ and row $v$ in $\mathbb{R}^n$, $A = uv$.

Question

Let $A$ be an $m \times n$ matrix with $\text{rank}(A) = 1$. Prove that for some nonzero column vector $u$ in $\mathbb{R}^m$ and a nonzero row vector $v$ in $\mathbb{R}^n$, the matrix $A$ can be written in terms of the outer product \begin{equation} A = uv. \end{equation}

I do not know where to start, can someone please guide me in the right direction?

user144410 · Answer 1 · 2017-03-12T16:01:36.563

Hint: $A$ is rank $1$ if and only if all its eigenvalues are equal to zero except 1. That is for any vector $u$ in $\mathbb{R}^m$, we have \begin{equation} u = \frac{1}{\lambda} A\tilde{v} = A v \end{equation} in which $\lambda$ is the eigenvalue of $A$. This is true for unit vectors (all entries are zero except one entry is 1). Then you can show that $A = v u^\top$

Next step you need to show that if $A = u v^\top$ then it must be rank $1$. To do this you need to show that for any vector $v$ in $\mathbb{R}^n$, $Av$ is a scalar multiple of $u$

Let $A$ be an $m\times n$ matrix with $\text{rank}(A) = 1$. Prove that for some column $u$ in $\mathbb{R}^m$ and row $v$ in $\mathbb{R}^n$, $A = uv$.

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