Linear algebra: Considering an affine map and formulating a conjecture Consider an affine map $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, $f(x) = Mx + s$ and let $S$ be a square. Formulate, and prove, a conjecture that relates how $\text{area}(f(S))$ depends on $\text{area}(S)$ in terms of parameters $M$ and $s$ of $f$.
Recommendations on how I should go about this?
 A: Very explicit hint:
Let's do it in a simpler case.
$f: \mathbb{R} \to \mathbb{R}$ affine map ($f(x)=ax+b$), same question.
$$\text{Vol}(f(S))=\int_{f(S)} 1 = \int_S |f'| = |a|\int_S 1  = |a|\text{Vol}(S) $$
A: We describe the square $ABCD$ by a position vector to a corner $p=OA$ and two side vectors $a=AB$ and $b=AD$ each of length $L$.
Then
$$
\DeclareMathOperator{area}{area}
\area(S) = \det(a,b) \\
S = \{ x = p + u a + v b \mid u,v \in [0, 1] \} \quad (*)
$$
and
\begin{align}
f(S) 
&= \{ f(x) \mid x \in S \} \\
&= \{ f(p + u a + v b) \mid u, v \in [0, 1] \} \\
&= \{ M(p + u a + v b) + s \mid u, v \in [0,1] \} \\
&= \{ (Mp+s) + u (Ma) + v (Mb) \mid u,v \in [0,1] \} \\
&= \{ p' + u a' + v b' \mid u, v \in [0,1] \} 
\end{align}
we notice that $f(S)$ is of form $(*)$.
Up to here it plays no role if $M$ is a scalar or a $2\times 2$ matrix.
The $S$ in $(*)$ in general is a parallelogram. We need the additional condition $a \perp b$ to make it a rectangle and 
$\lVert a \rVert = \lVert b \rVert$ to get a square.
$S' = f(S)$ is a parallelogram with position vector $p' = M p + s$ and sides $a'= M a$ and $b' = M b$. Depending on the matrix $M$ it might degenerate to a line segment or a single point.
For scalar $M$ we would have
$$
\area(f(S))
= \det(M a, M b) = M^2 \det(a, b) = M^2 \area(S)
$$
For the matrix case it would be
\begin{align}
\area(f(S)) 
&= \det(M a, M b) \\
&= (m_{11} a_1 + m_{12} a_2)(m_{21} b_1 + m_{22} b_2) 
- (m_{21} a_1 + m_{22} a_2)(m_{11} b_1 + m_{12} b_2) \\
&= (m_{11} m_{22} - m_{21} m_{12}) a_1 b_2 +
 (m_{12} m_{21} - m_{22} m_{11}) a_2 b_1 \\
&= \det M \det(a,b) \\
&= \det M \area(S)
\end{align}
