rank(A) $≤ 2 \iff A = U.V^T$

Question

Let $A \in M_{n×n}(\mathbb R)$, with $n ≥ 2$ a positive integer. Show that the following statements are equivalent :

i) rank(A) $≤ 2$.

ii) $\exists U,V ∈ M_{n×2}(\mathbb R)$ such that $A = U.V^T$

I honestly struggled to show both ways of the equivalence but I have a few guesses (I had the idea for ii $\implies$ i while writing this):

i $\implies$ ii : If rank(A) $≤ 2$ then there are $0,1$ or $2$ independent columns in A. We want to write A as a product of two nx2 matrixes. (I tried to take 4x4 matrixes of rank 2 and tried to write them in the form we are asking. I took U to be the independent columns, and V formed by the image of the canonic basis vectors and by pure luck it worked but this is not a proof).

ii $\implies$ i : We know that $rank(A)=rank(U.V^T) ≤ min(rank(U), rank(V^T))$, U and $V^T$'s ranks are clearly $\leq 2$ so rank(A) $\leq 2$

Can someone help me to write a solid proof for i $\implies$ ii instead of a lucky one ?

Answer 1

I think the easiest way to do that exercise is to interpret matrices as linear maps instead of something like two-dimensional arrays that can be added and multiplied.
In that case the statements become:

1. $\operatorname{rank}(A)\leq 2$, i.e. the image of $A$ has dimension at most $2$.
2. $A:\Bbb R^n\to\Bbb R^n$ factors as $\Bbb R^n\xrightarrow{V^T}\Bbb R^2\xrightarrow{U}\Bbb R^n$.

Now (I would say) the proof becomes simpler. First it is clear, that if $A$ factors as in $2$ then the image of $A$ has dimension at most $2$ (because it is contained in the image of $U$) (this is basically what you already have).
For the other direction assume that $\operatorname{rank}A=2$ (the other cases are similar, we just need fewer basis vectors). Choose a basis $v_1,v_2$ of the image of $A$. Let $\Bbb R^n\to \operatorname{im}A$ be the restricion of $A$. Composing this with the isomorphism $\operatorname{im}A\to \Bbb R^2$, $v_i\mapsto e_i$ will get us the map $V^T$. Now $U$ is just the map $e_i\mapsto v_i$. We see that $$\operatorname{im}A\to\Bbb R^2\xrightarrow{U}\Bbb R^n$$ is just the inclusion of the image of $A$ into $\Bbb R^n$. So in total the composition of $U$ and $V^T$ is just $A$ (as linear maps). Now if you represent $U,V^T$ in terms of the standard basis we get matrices that satisfy $A=UV^T$ (As matrix multiplication)

(Here I wrote $A,U,V^T$ for both the matrices and their linear maps)

Answer 2

Hint: Via Gauss's elimination, there exists $P,Q\in M_{n\times n}(\mathbb R)$ such that

$$P\begin{bmatrix}I_r&O\\O&O\end{bmatrix}Q=A$$

where $r=\operatorname{rank}(A)$.

When $r=2$, $A=\left(P\begin{bmatrix}I_r\\O\end{bmatrix}\right)\cdot\left(\begin{bmatrix}I_r&O\end{bmatrix}Q\right)$, implying ii) as desired.