Find the dimension of the image of the linear transformation I need to find the dimension of the image of the linear transformation $f(v)$, where $f\colon \Bbb R^2 \to \Bbb R^2$ is defined by
$$f(v) = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}v$$
I have already found that the linear transformation is neither injective nor surjective by finding contradictory examples for both, but I'm not sure where to proceed. I know that the $\text{Im} (f) = \{f(v) \} v$ elements of vector space $V$}, but I'm not sure how to derive the dimension.
 A: $$f(\left(
\begin{array}{c}
v_1\\
v_2
\end{array}\right))=\left(
\begin{array}{cc}1&0\\
0&0\end{array}\right)\cdot\left(
\begin{array}{c}
v_1\\
v_2
\end{array}\right)=\left(
\begin{array}{c}
v_1\\
0
\end{array}\right)$$
Hence the null-space of $f$ is
$$N(f)= \left\langle \left(
\begin{array}{c}
0\\
1
\end{array}\right) \right\rangle$$
then use  
$$\dim(N(f))+\dim(R(f))=\dim(\mathbb{R}^2)$$  
so that  
$$\dim(R(f))=\dim(\mathbb{R}^2)-\dim(N(f))=2-1=1$$.
A: $\dim(f)=\dim([1,0;0,0])$ because matrix of $f$ in standard basis is exactly $[1,0;0,0]$
A: Another perspective:  the image of the linear transformation that sends $\vec{x}$ to $A \vec{x}$ for some matrix $A$ is the column space of $A$, the subspace spanned by the columns of $A$.
In your example the columns of $A$ are $(1,0)$ and $(0,0)$.  Given a finite set $S = \{\vec{v}_1,\dots,\vec{v}_k$ of vectors that span a vector space $V$, you can delete vectors from $S$ to obtain a basis in the following manner.  Drop a vector $\vec{v}_i$ from the list if it is a linear combination of $\vec{v}_1,\dots, \vec{v}_{i-1}$, and keep it otherwise.  The vectors that you keep will be a basis for $V$.  (It is a good exercise to prove this statement.)
In your example, you would keep $\vec{v}_1 = (1,0)$.  (In the algorithm, you always keep the first vector $\vec{v}_1$ for your basis.)  Your second vector $\vec{v}_2 = (0,0)$ is a linear combination of $\vec{v}_1$, namely $\vec{v}_2 = 0 \vec{v}_1$, so you drop it.  Your basis consists of just the single vector $\vec{v}_1 = (1,0)$, so you conclude that the image of your transformation is one-dimensional and in fact the image is simply the span of the vector $(1,0)$.
