If for every $v$ there exists $w$ with $Aw = v$, then $A$ is invertible 
Let ${A}$ be a $2 \times 2$ matrix. For every two-dimensional vector ${v}$, there exists a two-dimensional vector ${w}$ such that
  $Aw = v$.
  Show that ${A}$ is invertible.

What is the easiest way to do this?
Thanks.
 A: Take $v_1 = (1, 0)^T$ and $v_2 = (0, 1)^T$, by condition, there exist $w_1, w_2$ such that $Aw_1 = v_1$ and $Aw_2 = v_2$. That is,
$$A[w_1, w_2] = [v_1, v_2] = I.$$
Thus $B = [w_1, w_2]$ satisfies $AB = I$, hence $A$ is invertible.
A: Let $\mathbf{w_1}$ satisfies $$A\mathbf{w_1}=\left(\array{1\\0}\right) $$ and  $\mathbf{w_2}$ satisfies $$A\mathbf{w_2}=\left(\array{0\\1}\right) $$
Then $$A(\mathbf{w_1}\mathbf{w_2}) = \left(\array{1&0\\0&1}\right)=I $$ 
A: I think you would like to see a proof of the following result: 

Let ${A}$ be a $2 \times 2$ matrix. For every two-dimensional vector ${v}$, there exists a two-dimensional vector ${w}$ such that
  $Aw = v$.
  Show that $\det (A)\neq 0$.

One of the  easiest ways to prove this result is by contradiction. 
Let 
$$ A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ 
Suppose for every two-dimensional vector ${v}$, there exists a two-dimensional vector ${w}$ such that $Aw = v$ and $\det (A)=ad-bc=0$.
Case 1: If a=b=c=d=0 then clearly there is no two-dimensional vector ${w}$ such that $Aw = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Contradiction. 
Case 2: If $\begin{pmatrix} a \\ c \end{pmatrix}\neq 0$. Then let  $\begin{pmatrix} x \\ y \end{pmatrix}$ be a two-dimensional vector such that $ A \begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -c \\ a \end{pmatrix}$. That is 
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -c \\ a \end{pmatrix}$$
So we have $$  \left \{ \begin{aligned} ax+by&=-c\\cx+dy&=a \end{aligned} \right. $$ Multiplying the first equation by $-c$, the second equation by $a$ and adding them, we get
$$ (ad-bc)y=a^2+c^2$$ 
But $ad-bc=\det (A)=0$. So we have $ a^2+c^2=0$. Contradiction, because  $\begin{pmatrix} a \\ c \end{pmatrix}\neq 0$.
Case 3: If $\begin{pmatrix} b \\ d \end{pmatrix}\neq 0$. It is completely analogous to case 2.
Remark: Note that this proof is completely elementar in the sense it does not require any knowledge about the properties of determinants nor the inverse of linear transformations.
