Finding the span of a subspace if I have its image as a matrix This is at extract from a written solution to a problem:

$$
U_1 =  \operatorname{Im} \begin{bmatrix}
2\cos^2 (\theta/2) & 2\sin(\theta/2)\cos(\theta/2)\\
2\sin(\theta/2)cos(\theta/2) & 2\sin^2 (\theta/2)
 \end{bmatrix}  \Rightarrow
$$
the columns are linear dependent and $ U_1 = \operatorname{span}\left( \begin{bmatrix} \cos(\theta/2) \\ \sin(\theta/2)\end{bmatrix}\right) $

I can see that the a vector that is the base of the span is a multiplier of that image. But why was it concluded that this is the span?
 A: Recall that for a linear transformation $T:V\to W$, if $B_V=\{v_1,v_2,...,v_n\}$ is a basis of $V$, then $\text{Im}(T)$ is spanned by $\{T(v_1),T(v_2),...,T(v_n)\}$.
$\text{rank}(U_1)=1\implies \text{Im}(U_1)$ has dimension $1$, and its basis is singleton. So the basis is either $\{T((1,0))\}$ or $\{T((0,1))\}$; that is, it is spanned by any one of the first and second columns of $U_1$. 
$\therefore\text{Im}(U_1)=\begin{cases}\text{span}\begin{bmatrix}2\cos^2(\theta/2)\\2\sin(\theta/2)\cos(\theta/2)\end{bmatrix},&\sin(\theta/2)=0\\\text{span}\begin{bmatrix}2\sin(\theta/2)\cos(\theta/2)\\2\sin^2(\theta/2)\end{bmatrix},&\cos(\theta/2)=0\end{cases}=\text{span}\begin{bmatrix}\cos(\theta/2)\\\sin(\theta/2)\end{bmatrix}$
A: The image of a matrix is the span of its columns. For example, if 2-by-2 matrix $A$ has columns $\overrightarrow{a}_1$ and $\overrightarrow{a}_2$, then 
$$
[\overrightarrow{a}_1 \overrightarrow{a}_2]\left[\begin{array}{c}x\\y\end{array}\right] = x \overrightarrow{a}_1 + y \overrightarrow{a}_2
$$
Since the image is the set of all outputs, $x$ and $y$ run through $\mathbb{R}$, so you get the span of the columns. A basis for the image is given by a maximal linearly independent set of columns. 
