Show that the determinant $\mathcal{D}(\mathbf{x}, \mathbf{y})$ is the signed area of the parallelogram spanned by $\mathbf{x}$ and $\mathbf{y}$.

Question

Full question (which I couldn't put in the title due to text limits):

I've done 4a. and 4b. with manual calculation since they're just $2\times2$ matrices.

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Here's my attempt on 4c.:

Four properties of the determinant $\mathcal{D}: \mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}$:

1. $\mathcal{D}(\mathbf{x}, \mathbf{y}) = -\mathcal{D}(\mathbf{x}, \mathbf{y})$
2. $\mathcal{D}(c\mathbf{x}, \mathbf{y}) = c\mathcal{D}(\mathbf{x}, \mathbf{y}) = \mathcal{D}(\mathbf{x}, c\mathbf{y})$ for $c\in\mathbb{R}$
3. $\mathcal{D}(\mathbf{x} + \mathbf{y}, \mathbf{z}) = \mathcal{D}(\mathbf{x}, \mathbf{z}) + \mathcal{D}(\mathbf{y}, \mathbf{z})$ and $\mathcal{D}(\mathbf{x}, \mathbf{y}+\mathbf{z}) = \mathcal{D}(\mathbf{x}, \mathbf{y}) + \mathcal{D}(\mathbf{x}, \mathbf{z})$
4. For standard basis vectors $\mathbf{e}_1, \mathbf{e}_2$, $\mathcal{D}(\mathbf{e_1}, \mathbf{e_2}) = 1$

Using the properties of the determinants, I'll show that the properties of the determinant imply that $\mathcal{D}(\mathbf{x}, \mathbf{y})=x_1y_2 - x_2y_1$ for $\mathbf{x}, \mathbf{y}\in\mathbb{R}^2$:

$\begin{align*} \mathcal{D}(\mathbf{x}, \mathbf{y})&=\mathcal{D}(x_1\mathbf{e}_1+x_2\mathbf{e}_2, y_1\mathbf{e}_1+y_2\mathbf{e}_2)\\ &=\underbrace{\mathcal{D}(x_1\mathbf{e}_1, y_1\mathbf{e}_1+y_2\mathbf{e}_2)}_a+\underbrace{\mathcal{D}(x_2\mathbf{e}_2, y_1\mathbf{e}_1+y_2\mathbf{e}_2)}_b \tag{by property 3} \\ \end{align*}$

Let's first evaluate $a$.

$\begin{align*} \mathcal{D}(x_1\mathbf{e}_1, y_1\mathbf{e}_1+y_2\mathbf{e}_2)&=\mathcal{D}(x_1\mathbf{e}_1, y_1\mathbf{e}_1)+\mathcal{D}(x_1\mathbf{e}_1, y_2\mathbf{e}_2) \tag{by property 3}\\ &= x_1\mathcal{D}(\mathbf{e}_1, y_1\mathbf{e}_1)+\mathcal{D}(x_1\mathbf{e}_1, y_2\mathbf{e}_2)\tag{by property 2}\\ &= x_1y_1\mathcal{D}(\mathbf{e}_1, \mathbf{e}_1)+\mathcal{D}(x_1\mathbf{e}_1, y_2\mathbf{e}_2)\tag{by property 2}\\ \end{align*}$

By property 1, $\mathcal{D}(\mathbf{e}_1, \mathbf{e}_1) =-\mathcal{D}(\mathbf{e}_1, \mathbf{e}_1)$ implies that $\mathcal{D}(\mathbf{e}_1, \mathbf{e}_1) = 0$. So,

$\begin{align*} x_1y_1\mathcal{D}(\mathbf{e}_1, \mathbf{e}_1)+\mathcal{D}(x_1\mathbf{e}_1, y_2\mathbf{e}_2)&=\mathcal{D}(x_1\mathbf{e}_1, y_2\mathbf{e}_2)\\ &=x_1\mathcal{D}(\mathbf{e}_1, y_2\mathbf{e}_2)\tag{by property 2}\\ &=x_1y_2\mathcal{D}(\mathbf{e}_1, \mathbf{e}_2)\tag{by property 2}\\ &=x_1y_2\tag{by property 4} \end{align*}$

Similarly for $b$,

$\begin{align*} \mathcal{D}(x_2\mathbf{e}_2,y_1\mathbf{e}_1+y_2\mathbf{e}_2)&=\mathcal{D}(x_2\mathbf{e}_2,y_1\mathbf{e}_1)+\mathcal{D}(x_2\mathbf{e}_2, y_2\mathbf{e}_2)\tag{by property 3}\\ &=\mathcal{D}(x_2\mathbf{e}_2, y_1\mathbf{e}_1)+0 \tag{shown with a}\\ &=x_2y_1\mathcal{D}(\mathbf{e}_2, \mathbf{e}_1) \tag{by property 2}\\ &=-x_2y_1\mathcal{D}(\mathbf{e}_1, \mathbf{e}_2) \tag{by property 1}\\ &=-x_2y_1 \tag{by property 4} \end{align*}$

And so,

$$\mathcal{D}(\mathbf{x}, \mathbf{y}) = a + b = x_1y_2 + (-x_2y_1)$$

Boom! Now we got proof that the properties of the determinants mean that in $\mathbb{R}^2$, $\mathcal{D}(\mathbf{x}, \mathbf{y})=x_1y_2 - x_2y_1$.

Because in the earlier parts of the text I was shown how to get from the area of a parallelogram to the determinant, I can in fact just "work backwards" to get the area of the parallelogram back:

$\begin{align*} \mathcal{D}(\mathbf{x}, \mathbf{y})&=x_1y_2 - x_2y_1\\ &= \rho(\mathbf{x})\cdot\mathbf{y}\\ &=||x||\hspace{1mm}||y||\cos(\frac{\pi}{2}-\theta) \tag{by definition of dot product}\\ &=||x||\hspace{1mm}||y||\sin(\theta) \end{align*}$

Here $\theta$ is the angle between the "un-rotated" vector $\mathbf{x}$ and $\mathbf{y}$ and $\frac{\pi}{2}-\theta$ is the angle the rotated vector $\rho(\mathbf{x})$ and $\mathbf{y}$.

Questions:

1. Is my proof correct? I think it must be wrong somewhere because I didn't use 4b. ($\text{det}(A_\theta B) = \det(B)$) here.
2. How would I use 4b. ($\text{det}(A_\theta B) = \det(B)$) in my proof?

Answer 1

Hmm, this looks mighty familiar to me. This is certainly not the approach the author intended. You're just deriving the formula for the determinant in the $2\times 2$ case from the defining properties, but that wasn't the point.

Here's how you should start. Choose $\theta$ so that $A_\theta\mathbf x$ lies on the positive $x$-axis. We know from part b. that $\mathcal D(\mathbf x,\mathbf y) = \mathcal D(A_\theta\mathbf x,A_\theta\mathbf y)$. But how do we see from the latter that we have the usual "base times height" for the area of a parallelogram?