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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!

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    $\begingroup$ Probably not what you are looking for.... definitely relevant none the less:math.stackexchange.com/questions/250534/… $\endgroup$ – Squirtle Dec 3 '13 at 4:48
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    $\begingroup$ Because the expansion is not unique, I would guess there is probably no natural interpretation. $\endgroup$ – Squirtle Dec 3 '13 at 5:38
  • $\begingroup$ @Squirtle What do you mean by not unique? $\endgroup$ – dfg Dec 3 '13 at 23:02
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    $\begingroup$ @dfg He just means that you can expand along any row or column, so the physical interpretation won't be unique. We can interpret the determinant as giving the volume of the parallelepiped spanned by the rows (as opposed to the columns). There are just multiple (essentially equivalent) ways to interpret the determinant geometrically. $\endgroup$ – D Wiggles Dec 3 '13 at 23:23
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    $\begingroup$ @IBWiglin perhaps even one interpretation is what the OP wants $\endgroup$ – Squirtle Dec 5 '13 at 0:40
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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».

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  • $\begingroup$ What does "$\Lambda^{\text{top}-1}V$" mean? $\endgroup$ – dfg Dec 5 '13 at 20:06
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    $\begingroup$ $(n-1)$'th exterior power of $V$, if $V$ is $n$-dimensional; if you don't know about exterior powers it's not really important for this answer (but in general very useful for understanding determinants, IMO). $\endgroup$ – Grigory M Dec 5 '13 at 20:09
  • $\begingroup$ I'm sorry, I'm completely lost. I followed you to the point where you derived the volume of a parallelepiped, but I don't see how that connects to the cofactor theorem at all! Could you maybe make the connection more explicit? $\endgroup$ – dfg Dec 5 '13 at 20:17
  • $\begingroup$ I understand the connection for a 3x3 matrix, its just the generalization to higher dimensions that confused me. $\endgroup$ – dfg Dec 5 '13 at 20:28
  • $\begingroup$ Just to be sure: the reason it works for a 3x3 case is because it simply computes the cross product and dot product right? $\endgroup$ – dfg Dec 5 '13 at 20:45
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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.

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  • $\begingroup$ What does $∧$ mean? $\endgroup$ – dfg Dec 7 '13 at 15:45
  • $\begingroup$ @dfg $\wedge$ is exterior product $\endgroup$ – Shuchang Dec 7 '13 at 15:47
  • $\begingroup$ @ShuchangZhang If $\wedge$ is exterior product, what is the meaning of $e_1\wedge\cdots\wedge e_n=e_1\cdots e_n$? How do you multiply elements of a vector space in RHS? $\endgroup$ – Grigory M Dec 7 '13 at 16:48
  • $\begingroup$ @GrigoryM $e_1\cdots e_n$ is geometric product and the result is $n$-vector(or $n$-blade) $\endgroup$ – Shuchang Dec 7 '13 at 16:50
  • $\begingroup$ Ah, so you work in the Clifford algebra essentially, I see. $\endgroup$ – Grigory M Dec 7 '13 at 16:57

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