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I've been trying to work through the exercises in my book where you have to prove certain things are the case. It has been going okay so far, however I've gotten stuck on the following exercise, and I am not sure where to start.

How can you prove that an $m \times n$ matrix $A$ has rank 1, if and only if $A$ can be written as the outer product uv$^T$ of a vector u in $\mathbb{R}^m$ and v in $\mathbb{R}^n$

I really don't know how to even start with this.

Mosbas
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    It's an if and only if statement, so you need to prove the two directions. Did you try any of them? – 5xum Feb 23 '16 at 10:13
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    You need to require both $u$ and $v$ to be non-zero, otherwise the outer product could be the zero matrix which does not have rank one. – Michael Albanese Feb 23 '16 at 10:14
  • @5xum I don't really understand what you mean by that? – Mosbas Feb 23 '16 at 10:18
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    @Mosbas You need to prove two statements: One is: "If $A=uv^T$ for $u\in\mathbb R^m,, v\in\mathbb R^n$, then $A$ has rank $1$." The other is: "If $A$ has rank $1$, then $A=uv^T$ for some pair of vectors $u,v$. Did you try proving at least one of these two statements? (I advise you first attempt the first one, it's easier). – 5xum Feb 23 '16 at 10:21
  • @5xum I have attempted it, but I am absolutely useless at this. At the moment I feel really frustrated because I've been trying this for ages and not getting anywhere. I have to leave now so I don't have the time to show what I attempted, I appreciate the help though. – Mosbas Feb 23 '16 at 10:26
  • Please post your exercise. – Ziyuan Feb 23 '16 at 11:29

2 Answers2

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Since the matrix $A$ has rank 1, the space of image of $\Bbb R^n$ under $A$ is one dimensional. Thus there exists a vector $u\in\Bbb R^m$ such that $||u||=1$ and $\mathcal R(A)=\text{span}(u)$. Observe that $f:\Bbb R^n\to \Bbb R$ defined by $$ f(w)=\alpha ;\ \ \ \text{where}\ \ \ \ Aw=\alpha u $$ is a linear functional. By Riesz Representation Theorem, there exists $v\in \Bbb R^n$ such that $$ f(w)=\langle w,v \rangle=v^{T}w $$. Therefore $\exists u\in\Bbb R^m$ and $\exists v\in\Bbb R^n$ such that $$ Aw=\alpha u = f(w)u=(v^{T}w)u=(uv^{T})w\ \ \ \forall w\in \Bbb R^n $$ which implies that $$ A=uv^{T} $$ for some $u\in\Bbb R^m$, $v\in\Bbb R^n$.

The converse is easy.

BigbearZzz
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$A$ has rank $1$ $\implies$ $A = uv^T$:

Rank $1$ means that $A$ is not the zero matrix, but all columns are proportional to one another. Let $u$ be the first non-zero column, and let $v$ consist of the coefficients of proportionality between $u$ and the other columns. For instance, if $$ A = \begin{bmatrix}0&2&3\\0&-3&-4.5\\0&6&9\end{bmatrix} $$ we would have $u = \left[\begin{smallmatrix}2\\-3\\6\end{smallmatrix}\right] $ and $v = \left[\begin{smallmatrix}0\\1\\1.5\end{smallmatrix}\right]$.

$A = uv^T$ $\implies$ $A$ has rank $1$:

$u$ is an $m \times 1$ matrix with rank $1$, and $v$ is an $n\times 1$ matrix with rank $1$, so the rank of their product must be $1$.

Arthur
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