# How to motivate the surface element


But my question is:

Is there an intuitive way to see that the Gramian $\det G(t)$ is the squared volume of said parallelotope?

• Do you know a proof already? Is that proof not "intuitive" to you? I know a proof, and it seems pretty intuitive to me, but intuitions are different for different people. Also, this may be more suitable for math.stackexchange since it is really more about math content than teaching (even though this content is being taught). – Steven Gubkin Jan 12 '16 at 17:04
• @StevenGubkin That's a good point indeed. At the time of asking I did not know a proof. Well, put differently, the books I used had the Gramian as the definition of said volume. Hence, I did not ask for a proof but for an intuitive explanation. Now I have some, and will post is as an answer. Probably you could then add yours. – Dirk Jan 18 '16 at 10:11
• It is probably worth pointing out to the students that this gives the right values when $k=1$ (the Pythagorean theorem) and $k=n$ (the determinant). – David E Speyer Jan 21 '16 at 2:49
• @davidspyer Sure! That's what I did so far, but I found it unclear why the intermediate cases look so different. – Dirk Jan 21 '16 at 15:50

Here is an argument that has convinced (or at least silenced :-)) students in office hours. Making it into a rigorous proof would require more time talking about what volume MEANS than I do.

First, suppose I have $k$ vectors $\vec{v}_1$, $\vec{v}_2$, ..., $\vec{v}_k$ in $\mathbb{R}^k$. Let $A$ be the $k \times k$ matrix with columns $\vec{v}_1$, ..., $\vec{v}_k$. Then the volume of the parallelotope is $$|\det(\vec{v}_1, \ldots, \vec{v}_k)| = |\det A| = \sqrt{\det(A^T A)} = \sqrt{\det(\vec{v}_i \cdot \vec{v}_j)}.$$

The nifty thing is that each entry $\vec{v}_i \cdot \vec{v}_j$ of the $k \times k$ matrix is invariant under rotation of coordinates. So we have expressed the rotationally invariant volume in terms of much simpler rotationally invariant quantities.

If the vector $\vec{v}_1$, ..., $\vec{v}_k$ are now in $\mathbb{R}^n$ for some $n>k$, it is still true that the volume is $\sqrt{\det(\vec{v}_i \cdot \vec{v}_j)}$, since we can just choose orthogonal coordinates in the $k$-plane spanned by the $\vec{v}$'s and ignore the larger ambient space. (On one occasion, I drew two vectors on a notepad, shaded the parallogram and tilted the notepad at various angles, to point out that the area and the dot products don't care how I hold the notepad in space.) And we can still write this as $\sqrt{\det(A^T A)}$.

• For someone who really understands QR decomposition, this is the same as Dirk's answer, but I claim it is more friendly to those who don't. – David E Speyer Jan 21 '16 at 20:30
• I wonder whether it might be useful to spell out the $k=2$ case: $\det \left( \begin{smallmatrix} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} \end{smallmatrix} \right) = |u|^2 |v|^2 - |u|^2 |v|^2 \cos^2 \theta = |u|^2 |v|^2 \sin^2 \theta$. Sometimes I find that students whose eyes are glazing over in linear algebra wake up when I remind them of high school geometry/trig. (And sometimes I learn that they never had high school geometry, which is a different problem.) – David E Speyer Jan 21 '16 at 21:44
• That is nice. I wonder if their is a similarly "geometric" proof of $k=2$ – Steven Gubkin Jan 22 '16 at 4:20
• Indeed, emphasizing the invariance of the expression seems a good idea. There are some students who are convinced by just that. – Dirk Jan 22 '16 at 10:10


Let's assume that you know that the volume of an $n$-dimensional parallelotope $A$ spanned by $n$ vectors $a_1,\dots,a_n\in\RR^n$ is $\vol_n(A) = \det(a_1,\dots,a_n)$.

To get the $k$-dimensional volume $\vol_k(A)$ of the parallelotope $A$ spanned by $a_1,\dots,a_k\in\RR^n$ (still in $\RR^n$!) do the following: Do $QR$-factorization of $A=[a_1,\dots,a_k]$ to get $$A = QR$$ with $Q\in\RR^{n\times n}$ and $R\in\RR^{n\times k}$ upper triangular. This is nothing else than turning (and mirroring) the parallelotope around by $Q^T$ such that $$\begin{array}{cl} r_1 = Q^T a_1&\in\span(e_1)\\ r_2 = Q^Ta_2&\in\span(e_1,e_2)\\ \vdots & \vdots\\ r_k=Q^Ta_k&\in \span(e_1,\dots,e_k). \end{array}$$ Since $Q^T$ is an isometry, we have that the $k$-dimensional volume of the parallelotope spanned by $a_1,\dots,a_k$ is the same as the one spanned by the columns of $R$. These lie in $\RR^k$ anyway, and hence, this volume is the determinant of the upper $k\times k$ part of $R$, i.e. the determinant of the diagonal entries of $R$. In other words: We have $$R = \left[\begin{matrix}R_1\\ 0\end{matrix}\right].$$ and the volume is $\vol_k(A) = \det(R_1)$.

Finally, we see that $$\det(A^TA) = \det(R^TQ^TQR) = \det(R^T I R) = \det(R_1^T R_1) = \det(R_1)^2 = \vol_k(A)^2.$$

My proof is basically the same as yours, but perhaps this will be a bit more intuitive.

We want to find the volume of a $k$-parallelepiped spanned by $v_1,v_2,...,v_k$ in $\mathbb{R}^k$. We are armed only with the determinant, which lets us find the (signed) volume of $m$-parallelepipeds in $\mathbb{R}^m$.

The crucial observation is that, for any good definition of volumes, we should have that the "$m$ -volume" of $v_1,v_2,...,v_m \in \mathbb{R}^n$ should be the same as the "$m+1$-volume" of $v_1,v_2,...,v_m,w \in \mathbb{R}^n$ if $w$ is perpendicular to all the $v_j$, and is of length $1$. In other words "Volume equals base times height".

So the volume of $v_1,v_2,...,v_k \in \mathbb{R}^n$ is the same as the volume of $v_1,v_2,...,v_k,w_1,w_2,...,w_{n-k}$, where each $w_i$ is orthogonal to all the other vectors in the list, and is of norm $1$.

So we could just say that the volume is $Det(v_1,v_2,...,v_k,w_1,w_2,...,w_{n-k})$. This is a recipe for how to find the volume, but it suffers some flaws as a definition and as a computational tool of $k$-volumes. As a definition, it is difficult because we would need show that the result is independent of the choice of $w_i$, and because the sign is meaningless (it depends on the $w_k$). As a computational tool, one actually needs to carry out the laborious process of calculating $w_i$ which work. It seems we should be able to find a formula which does not involve actually calculating these auxiliary vectors.

The insight is to realize that the determinant of the transpose is the same as the determinant of the original matrix. So letting $M = (v_1 v_2 ... v_k w_1 ... w_{n-k})$ and $A = (v_1 v_2 ... v_k)$

$\textrm{Vol}^2 = Det(M^\top M)$

But when we actually compute the matrix $M^\top M$, we see that it consists of $A^\top A$ and then ones down rest of the diagonal. So its determinant is just $Det(A^\top A)$.

Thus $\textrm{Vol}^2 = Det(A^\top A)$.

• I still have some feeling that perhaps, with a deeper understanding of adjoints and determinants, one could make this result "obvious" even without the auxiliary vectors. The magic is really happening in the step where we use that the determinant of the adjoint is the same as the determinant of the original map. – Steven Gubkin Jan 18 '16 at 17:06
• I like this argument. While it is similar from a computational point of view, it differs conceptually. I use that rotations should not change volume and you use "base times height" which also makes a lot of sense. And indeed my intuition is different from yours... – Dirk Jan 18 '16 at 21:22
• @Dirk, ya I only glanced at your argument and thought it was similar, but it really does have a different thought process behind it. – Steven Gubkin Jan 18 '16 at 22:05