If matrix $T$ has rank $1$ then it can be written as $T=\underline{uv}^T$ In my textbook there was the following statement:

If matrix $T$ has rank $1$ then it can be written as $T=\underline{{u}{v}}^T$.

I'm trying to figure out why it's true. For example:
$$
T=\begin{bmatrix}1 & 2 & 3\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
$$
The matrix $T$ has rank $1$ but how do I re-write it as $T=\underline{u}\underline{v}^T$? what is the general decomposition for any $T$?
 A: Take any nonzero column of the matrix as $u$. Since the rank is $1$, each column is a scalar multiple of this one.
Put these scalars as the coordinates of $v$.
A: We have that
$$T=\begin{bmatrix}1 & 2 & 3\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}=
\begin{bmatrix}1 \\
0 \\
0 
\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\
\end{bmatrix}$$
and more in general
$$T=\begin{bmatrix}ia & ib & ic\\
ja & jb & jc\\
ka & kb & kc
\end{bmatrix}=
\begin{bmatrix}i \\
j \\
k 
\end{bmatrix}\begin{bmatrix}a & b & c\\
\end{bmatrix}$$
A: If $T:V\to W$ has rank $1$, then its image has dimension $1$, meaning that it is spanned by a single vector $u$. So $Tx=\lambda(x) u$, where $\lambda(x)$ is a scalar depending on $x$. And since $T$ is linear, $\lambda:V\to K$ ($K$ being the field we're working over) also has to be linear. But then it has a matrix representation $\lambda (x)=\Lambda x$, where $\Lambda$ is an $n\times 1$ matrix, $n$ being the dimension of $V$. In other words, it's a row vector, so $\Lambda=v^T$ for some $v\in V$. Then we have
$$\lambda(x)=\Lambda x=v^Tx.$$
Putting everything together:
$$Tx=\lambda(x)u=u\lambda(x)=uv^Tx.$$
So in the end, $T=uv^T$, where $u$ is a basis vector of $\operatorname{im} T$, and $v^T$ is chosen such that $uv^Tx=Tx$ for all $x\in V$. To find its components, you have to calculate $Tb_i$ for every basis vector $b_i$ of $V$. If the basis is orthonormal (like the standard basis of $\mathbb R^n$), then
$$Tb_i=uv^Tb_i=u v_i,$$
which will allow you to obtain the $i$-th component $v_i$ of $v$ in the basis $\{b_i~\vert~i=1,\dots,\dim V\}$.
