Why Is the Result of the Convolution of a Row and a Column a Two Dimensional Matrix?

I'm trying to understand a book chapter on "Algorithms For Efficient Computation of Convolution" (Also on Doc Droid or Scribd) and I know that when calculating the convolution of an image represented by two dimensional matrix (array) with a mask matrix the result is the treated two dimensional image.

How to calculate the convolution of two signals represented by a row and a column?

The chapter I'm reading says that:

Separable convolution kernel must fullfil the condition that its matrix has rank equal to one. In other words, all the rows must be linearly dependent. Why? Let us construct such a kernel. Given one row vector

$$\vec{u}=\left(u_1, u_2, u_3, \ldots, u_m\right)$$ and one column vector $$\vec{v}^T=\left(v_1, v_2, v_3, \ldots, v_n\right)$$ let us convolve them together: $$\vec{u}*\vec{v}=\left(u_1, u_2, u_3, \ldots, u_m\right)*\begin{pmatrix}v_1\\v_2\\v_3\\\vdots\\v_n\end{pmatrix}=\begin{pmatrix}u_1v_1&u_2v_1&u_3v_1&\cdots&u_mv_1\\ u_1v_2&u_2v_2&u_3v_2&\cdots&u_mv_2\\ u_1v_3&u_2v_3&u_3v_3&\cdots&u_mv_3\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ u_1v_n&u_2v_n&u_3v_n&\cdots&u_mv_n\end{pmatrix}=A\tag{4}\\$$

I found that the second example was more relevant for your question. I want to give a $\begin{bmatrix}1&2 &1 \end{bmatrix}$ weight column wise and on top of that give a $\begin{bmatrix}1\\2 \\1 \end{bmatrix}$ weight to each row. The result is the filter $\begin{bmatrix}1&2 &1\\2&4&2 \\1&2 &1 \end{bmatrix}$ which take into consideration both your demands. By the way this spesific filter will smooth you image.