Linear algebra: Considering an affine map and formulating a conjecture

Question

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?

Answer 1

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}

Answer 2

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.

$$\text{Vol}(f(S))=\int_{f(S)} 1 = \int_S |f'| = |a|\int_S 1 = |a|\text{Vol}(S) $$