# Determinant of non-square matrices.

Suppose I have an linear transformation described by a square matrix $A$. Then, $A$ takes a vector $x\in\mathbb{R}^n$ to a vector $y\in\mathbb{R}^n$ related to $x$ by the equation $y=Ax$.

Grant Sanderson defined the determinant of $A$ on his youtube series on linear algebra in the following way:

Two vectors $x$ and $y$ determine a unique paralelogram. On the same way $Ax$ and $Ay$ determine another paralelogram. The ratio between its areas is defined to be the determinant of $A$. That is $|\det(A)|=\text{Area}(Ax,Ay)/\text{Area}(x,y)$. (The sign of the determinant is given by the orientation of the vectors. If $A$ preserves orientation, its determinant is positive.)

Can't we define a determinant to non-square matrices by the same formula?

EDIT: (Example) Take $$A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ and $$x=\begin{bmatrix} 1 \\ 0\end{bmatrix}, \quad y=\begin{bmatrix} 0 \\ 1\end{bmatrix}.$$ Using the notation defined above, we clearly have $\text{Area}(x,y)=1$. We have that $$Ax=\begin{bmatrix} 1 \\ 3\end{bmatrix}, \quad Ay=\begin{bmatrix} 2 \\ 4\end{bmatrix}$$ too. So, $\text{Area}(Ax,Ay)=2$. Then the ratio between the areas is $2$, which coincides with the modulus of the determinant of $A$. (Of course $A$ changes the orientation of the vectors, so it's determinant is negative.)
EDIT 2: Here's an explicit definition of the determinant. $$|\det(A)| = \begin{cases} \text{Area}(Ax,Ay)/\text{Area}(x,y) & \quad \text{if } \text{kernel}(A) \text{ is zero}\\ 0 & \quad \text{if } \text{kernel}(A) \text{ is non-zero}\\ \end{cases},$$ for any linearly independent vectors $x$ and $y$.
If $A$ preserves orientation, then $\det(A)>0$. $\det(A)\leq 0$ otherwise.
• @copper.hat But it coincides with the determinant when $A$ is a square matrix right? – Gabriel Aug 17 '16 at 1:53
• @GabrielRibeiro What about an $m\times n$ matrix with $n>m$? The determinant you defined is always zero? – velut luna Aug 17 '16 at 1:58
• @AlphaGo The quantity $\text{Area}(Ax,Ay)/\text{Area}(x,y)$ makes sense even if $n>m$. I don't see why that is a problem. – Gabriel Aug 17 '16 at 2:04