Solution
The short answer is that \(\text{rank}(A) = 2\) if and only if \(\vec u\) and \(\vec w\) are linearly independent and \(\vec v\) and \(\vec z\) are linearly independent. If either \(\lbrace \vec u, \vec w \rbrace\) or \(\lbrace \vec v, \vec z \rbrace\) are linearly dependent, then \(\text{rank}(A) = 1\). The easy way to reason about this is that:
The columns of \(\vec u \vec v^T\) are all scalar multiples of \(\vec u\) and the columns of \(\vec w \vec z^T\) are all scalar multiples of \(\vec w\). If \(\vec u\) and \(\vec w\) are linearly dependent, then all columns of \(A\) are scalar multiples of \(\vec u\) (or \(\vec w\)), meaning \(\text{dim}(\text{colsp}(A)) = 1\), which means \(\text{rank}(A) = 1\).
The rows of \(\vec u \vec v^T\) are all scalar multiples of \(\vec v^T\) and the rows of \(\vec w \vec z^T\) are all scalar multiples of \(\vec z^T\). If \(\vec v\) and \(\vec z\) are linearly dependent, then all rows of \(A\) are scalar multiples of \(\vec v^T\) (or \(\vec z^T\)), meaning \(\text{dim}(\text{rowsp}(A)) = 1\), which means \(\text{rank}(A) = 1\).
In order for \(\text{rank}(A) = 2\), there need to be at least two linearly independent columns and two linearly independent rows, which means that both pairs \(\lbrace \vec u, \vec v \rbrace\) and \(\lbrace \vec w, \vec z \rbrace\) must be linearly independent.
For a more detailed explanation, let’s start from the beginning.
We are given
$$ A = \vec{u}\vec{v}^T + \vec{w}\vec{z}^T $$
with each outer product being rank 1. As the hint suggests, let’s think about what happens when multiplying \(A\) by a vector \(\vec x \in \mathbb{R}^n\):
$$ A \vec x = (\vec{u}\vec{v}^T + \vec{w}\vec{z}^T) \vec x = (\vec{v}^T\vec{x})\vec{u} + (\vec{z}^T\vec{x})\vec{w} $$
This shows that \(A \vec x\) is always a linear combination of \(\vec{u}\) and \(\vec{w}\). So, \(\text{colsp}(A)\) is a subspace of \(\text{span}(\lbrace\vec{u}, \vec{w}\rbrace)\), since every vector in \(\text{colsp}(A)\) can be written as a linear combination of \(\vec{u}\) and \(\vec{w}\).
If \(\vec{u}\) and \(\vec{w}\) themselves point in the same direction (i.e. are linearly dependent), then any vector in \(\text{colsp}(A)\) is a scalar multiple of \(\vec u\) (or, equivalently, of \(\vec w\)). This would mean \(\text{dim}(\text{colsp}(A)) = 1\), which means \(\text{rank}(A) = 1\).
With that case out of the way, let’s suppose \(\vec u\) and \(\vec w\) are linearly independent. Does this automatically mean that \(\text{dim}(\text{colsp}(A)) = 2\)? No: all we know for sure is that every vector of the form \(A \vec x\) is a linear combination of \(\vec u\) and \(\vec w\). We don’t yet know that the converse is true, i.e. that every linear combination of \(\vec u\) and \(\vec w\) can be written as \(A \vec x\), for some \(\vec x \in \mathbb{R}^n\).
In order for it to be the case that \(\text{colsp}(A) = \text{span}(\lbrace\vec{u}, \vec{w}\rbrace)\), we must be able to take any linear combination of \(\vec{u}\) and \(\vec{w}\) and write it as \(A \vec x\) for some \(\vec x \in \mathbb{R}^n\). Meaning, for any \(c\) and \(d\), we must be able to find some \(\vec x\) such that:
$$ A \vec x = c \vec u + d \vec w $$
But, we know that
$$ A \vec x = (\vec{v}^T\vec{x})\vec{u} + (\vec{z}^T\vec{x})\vec{w} $$
So, the question really is, is it always possible, for any \(c\), \(d\), to find an \(\vec x\) that satisfies:
$$ c = \vec{v}^T\vec{x}, \qquad d = \vec{z}^T\vec{x} $$
This may seem a bit abstract, so let’s just plug in specific numbers for \(c\) and \(d\) to make things a bit more clear.
$$ 3 = \vec{v}^T\vec{x}, \qquad 4 = \vec{z}^T\vec{x} $$
Remember that \(\vec v\) and \(\vec z\) are fixed here. Given them, can we find an \(\vec x\) that satisfies both of these equations? We can, as long as \(\vec v\) and \(\vec z\) are linearly independent too! If they are, then \(\vec v^T \vec x = 3\) is one equation with \(n\) unknowns, and \(\vec z^T \vec x = 4\) is another equation with \(n\) unknowns. Since there are \(n\) unknowns and 2 equations, the system is overdetermined, and there are infinitely many solutions for \(\vec x\), we just need to pick one of them.
Again, let’s give an example. Suppose \(\vec u\) and \(\vec w\) are linearly independent (a prerequisite for the case we’re considering), let \(\vec v = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}\) and \(\vec z = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\) (which are also linearly independent as a pair), and suppose \(A \vec x = 3 \vec u + 4 \vec w\). Then, \(\vec x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\) must satisfy:
$$ \vec v^T \vec x = 3 \implies 3x_1 + 4x_2 + 5x_3 = 3 $$
$$ \vec z^T \vec x = 4 \implies 0x_1 + 0x_2 + 1x_3 = 4 $$
Solving these, we get:
$$ x_3 = 4 $$
$$ 3x_1 + 4x_2 + 20 = 3 \implies 3x_1 + 4x_2 = -17 $$
So, as long as we pick \(x_3 = 4\) and \(x_1\) and \(x_2\) to satisfy \(3x_1 + 4x_2 = -17\), we can find an \(\vec x\) that satisfies both of the equations, meaning we’re able to find an \(\vec x\) to make \(A \vec x = 3 \vec u + 4 \vec w\).
Nothing was special about \(c = 3\) and \(d = 4\). What was special was that in addition to \(\lbrace \vec u, \vec w \rbrace\) being linearly independent, we also had \(\lbrace \vec v, \vec z \rbrace\) being linearly independent. This is the condition that makes it possible to find an \(\vec x\) that satisfies both of the equations. If they weren’t linearly independent, it isn’t guaranteed that we can find an \(\vec x\) that satisfies both of the equations. Suppose we revisit the same example, but instead with \(\vec v = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}\) and \(\vec z = \begin{bmatrix} 30 \\ 40 \\ 50 \end{bmatrix}\) (which are linearly dependent as a pair). Then, \(\vec v^T \vec x = 3\) and \(\vec z^T \vec x = 4\) are two equations with \(n\) unknowns still, but \(\vec x\) would need to satisfy
$$ 3x_1 + 4x_2 + 5x_3 = 3 $$
$$ 30x_1 + 40x_2 + 50x_3 = 4 $$
which has no solutions, meaning we’re not able to find an \(\vec x\) to make \(A \vec x = 3 \vec u + 4 \vec w\), meaning \(\text{colsp}(A) \neq \text{span}(\lbrace\vec{u}, \vec{w}\rbrace)\).
That was a long solution, in which we tried to build intuition for how you might think about this. But to summarize:
\(\text{rank}(A) = 2\) if and only if \(\vec u\) and \(\vec w\) are linearly independent and \(\vec v\) and \(\vec z\) are linearly independent.
If either \(\lbrace \vec u, \vec w \rbrace\) or \(\lbrace \vec v, \vec z \rbrace\) are linearly dependent, then \(\text{rank}(A) = 1\).
If any of the vectors are the zero vector, then \(\text{rank}(A) = 0\).