Linear maps from $\mathbb{R}^{2}$ to $\mathbb{R}^{3}$ and cross products.

Question

Let $S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ be a linear map, and let $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be an orthogonal map. Let $\{ e_{1}, e_{2} \}$ be the standard basis of $\mathbb{R}^{2}$. Is it true that \begin{equation*} \| S(e_{1}) \times S(e_{2}) \| = \| ST(e_{1}) \times ST(e_{2}) \|? \end{equation*} If the rank of $S$ is $0$ or $1$, then both sides of the equation are equal to zero. So that leaves the case that the rank of $S$ is $2$, but I got pretty much stuck there. I feel like this should be true, because it feels like the quantity I'm trying to calculate is somehow a property of the linear map $S$, that shouldn't depend on the choice of basis, but maybe that's just wrong.

Answer 1

Yes, it is true. Suppose that the matrix of $S$ with respect to the canonical basis is$$\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}.$$Then$$S(e_1)\times S(e_2)=(a,c,e)\times(b,d,f).$$On the other hand, if the matrix of $T$ with respect to the canonical basis is$$\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$$then the matrix of $S\circ T$ with respect to the canonical basis is$$\begin{bmatrix}a \cos (\theta )+b \sin (\theta ) & b \cos (\theta )-a \sin (\theta ) \\ c \cos (\theta )+d \sin (\theta ) & d \cos (\theta )-c \sin (\theta ) \\ e \cos (\theta )+f \sin (\theta ) & f \cos (\theta )-e \sin (\theta )\end{bmatrix}.$$So,\begin{multline}\bigl\|S\bigl(T(e_1)\bigr)\times S\bigl(T(e_2)\bigr)\bigr\|=\\=\bigl((a,c,e)\cos(\theta)+(b,d,f)\sin(\theta)\bigr)\times\bigl(-(a,c,e)\sin(\theta)+(b,d,f)\cos(\theta)\bigr)=\\=(\cos^2\theta+\sin^2\theta)(a,c,e)\times(b,d,f)=(a,c,e)\times(b,d,f).\end{multline}If the matrix of $T$ with respect to the canonical basis is$$\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix},$$the computations are similar.