Geometric interpretation of the cofactor expansion theorem

I find the geometric interpretation of determinants to be really intuitive - they are the "area" created by the column vectors of the matrix.

Could someone give me a geometric interpretation of the cofactor expansion theorem using the definition of the determinant as the "area"?

Thanks!

Of course this theorem has a geometric interpretation! In a sense, it's a multidimensional analogue of «the volume of a parallelepiped is the product of the area of its base and its height».

3. Let's start with $3\times3$ case: $$ \left|\begin{matrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right|= u_1\left|\begin{matrix}v_2&v_3\\w_2&w_3\end{matrix}\right| -u_2\left|\begin{matrix}v_1&v_3\\w_1&w_3\end{matrix}\right| +u_3\left|\begin{matrix}v_1&v_2\\w_1&w_2\end{matrix}\right|. $$ LHS is the volume of the parallelepiped spanned by three vectors, $u$, $v$ and $w$. What's the meaning of RHS? Clearly that's a scalar product of $u$ with something — namely, with the vector $$ \left(\left|\begin{matrix}v_2&v_3\\w_2&w_3\end{matrix}\right|, -\left|\begin{matrix}v_1&v_3\\w_1&w_3\end{matrix}\right|,\left|\begin{matrix}v_1&v_2\\w_1&w_2\end{matrix}\right|\right)= \left|\begin{matrix}\overrightarrow{e_1}&\overrightarrow{e_2}&\overrightarrow{e_3}\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right| $$ — i.e. with vector product of $v$ and $w$.

So the formula we get is $vol\langle u,v,w\rangle=(u,[v,w])$; now by the (geometrical) definition of scalar product it's $area\langle v,w\rangle\cdot (|u|\cdot\sin\phi)$, and the first factor is the area of the base and the second one is the height of our parallelepiped.

n. Consider the (general) case of vectors in $n$-dimensional space $V$. In RHS of the theorem we again see a scalar product of the first vector, $v$, with a vector $B$ (in coordinate-free language it really lives in $\Lambda^{n-1}V$, but let's ignore this for now) with coordinates $C_{1i}$.

The question is, what is the geometric meaning of $B$. Let me give 3 (closely related) answers.

1. By the very same cofactor theorem it measures the [(n-1)-dimensional] area of projection of the base of our $n$-parallelepiped (i.e. $(n-1)$-parallelepiped spanned by all vectors but $v$) on different hyperplanes; more precisely, the area of the projection on the hyperplanes orthogonal to a unit vector $v$ is the scalar product $(B,v)$.
2. Let's prove the cofactor theorem instead of using it. The function $(B,x)$ is linear in $x$. For a basis vector $x=e_i$ we have $(B,x)=C_{1i}$, which (up to sign, at least) is the area of the span of projections of our vectors on the hyperplane orthogonal to $e_i$. So $(B,x)$ is indeed the area of the projection of the base on the hyperplane orthogonal to $x$ (multiplied by $|x|$ and taken with appropriate signs).
3. Even better, since everything is invariant under (special) orthogonal transforms, let's change basis to make $v$ a scalar multiple of $e_1$. Now the statement «$(B,v)$ is the $|v|$ times the area of the projection» became obvious (we literally multiply $|v|$ by the cofactor manifestly equal to this area — well, it was discussed in (2) anyway).

Now I must admit the statement we get is more like «the volume of a parallelepiped $\langle u,\text{base}\rangle$ is the product of the length of $u$ and the area of the projection of its base on the hyperplane orthogonal to $u$» — but it's of course equivalent to «the volume of a parallelepiped is the product of the area of its base and its height».

Let me explain using geometric algebra. Take an orthonormal basis $e_1,\ldots,e_n$ and let columns of $A$ be $a_1,\ldots,a_n$. Then the determinant of $A$ equals to the volume of parallelepipe spanned by columns of $A$. That is, $$a_1\wedge\cdots\wedge a_n=\det Ae_1\wedge\cdots\wedge e_n$$ Or we can write $$\det A=(e_1\wedge\cdots\wedge e_n)^{-1}(a_1\wedge\cdots\wedge a_n)$$ Since $e_i$ are orthonormal, $e_1\wedge\cdots\wedge e_n=e_1\cdots e_n$ and hence $$(e_1\wedge\cdots\wedge e_n)^{-1}=(e_1\cdots e_n)^{-1}=e_n\cdots e_1=e_n\wedge\cdots\wedge e_1$$ Note that $e_n\wedge\cdots\wedge e_1$ is a subspace of $a_1\wedge\cdots\wedge a_n$ , we can further write $$\begin{align}\det A&=(e_n\wedge\cdots\wedge e_1)\cdot(a_1\wedge\cdots\wedge a_n)\\&=(e_n\wedge\cdots\wedge e_2)\cdot(e_1\cdot(a_1\wedge\cdots\wedge a_n))\\ &=(e_n\wedge\cdots\wedge e_2)\cdot\Big(a_{11}(a_2\wedge\cdots\wedge a_n)-\sum_{i=2}^n(-1)^ia_{1i}(a_1\wedge\cdots\hat a_i\cdots\wedge a_n)\Big)\\ \end{align}$$

• The first line is certainly the geometric explanation of determinant as mentioned above.
• The second line is exactly the geometric explanation of Laplace expansion. For one thing, $e_1\cdot(a_1\wedge\cdots\wedge a_n)$ extracts the "height" of parallelepipe with base in subspace $e_n\wedge\cdots\wedge e_2$. Next multiplied by $e_n\wedge\cdots\wedge e_2$ to restore the volume of "compressed" parallelepipe.
• The third line shows how to do the "extract", i.e. through "extracting" each edge.

This view can even be generalizd. For example, we could write $$\det A=(e_n\wedge\cdots\wedge e_3)\Big((e_2\wedge e_1)\cdot(a_1\wedge\cdots\wedge a_n\Big)$$ That means we can "extract" the "projection area" onto $e_2\wedge e_1$ of parallelpipe with base in subspace $e_n\wedge\cdots\wedge e_3$ and then restore the volume of "compressed" parallelepipe.