# Why Is $\sqrt{\det(A^TA)}$ A Volume / Volume Factor?

The determinant of an $n\times n$ matrix is a volume / volume factor. So far, I'm good in my understanding. You take a linear map, encode it as a matrix, compute the volume of the parallelepiped (or whatever the proper name is) spanned by the column vectors, and look at the factor by which this transformation scaled the unit $n$-dimensional volume from before the transformation to the new one. That scaling is the determinant. There are many ways to view the determinant, but this is the most interesting to me, because I can visualize it.

Now, what if I have a transformation from $\mathbb{R}^n$ to $\mathbb{R^m}$, encode it by an $m\times n$ matrix, and want the $n$-dimensional volume of the parallelepiped spanned by the column vectors of my matrix? This is a well-grounded question (think of a 2-d parallellogram embedded arbitrarily in 3-space: what is it's area?), but pretty much never addressed in linear algebra courses / books. Apparently (check e.g. the Wikipedia entry for determinants) I'm supposed to compute $\sqrt{\det(A^TA)}$ now.

This makes sense in the $m=n$ scenario (except that orientation changes might be lost due to the square root?), since $|\det(A)|=\sqrt{\det(A)^2}=\sqrt{\det(A^T)\det(A)}=\sqrt{\det(A^TA)}$, but I just can't visualize it in the case $n\neq m$. I see that the end result of the product $A^TA$ is an $n\times n$ matrix, so clearly the determinant is then an n-dimensional volume / volume factor, but I can't see why I get the correct volume. Any help?

• Denote the root of the determinant by $D(A)$ and the volume by $V(A)$. If $P$, $Q$ are orthogonal matrices, then it's clear that $D(PAQ) = D(A)$. If it's also true that $V(PAQ) = V(A)$, you can use en.wikipedia.org/wiki/Singular_value_decomposition to reduce to the case of an $n$-cell with edges in the axes, which is easy to check directly. I tend to believe $V(PAQ) = V(A)$ but I'm not sure of a proof. Aug 4, 2016 at 14:01
• @JoshuaCiappara Well, considering that orthogonal matrices have determinant $\pm 1$ they certainly are volume preserving by the exact interpretation of the determinant I have stated in my question (and that $\det(AB)=\det(A)\det(B)$, but that is a direct consequence of the same interpretation of the determinant). I think everything checks out perfectly now, thank you! Aug 4, 2016 at 15:48
• Well, except that orthogonal matrices of size $k$ preserve $k$-volume, and $P$ and $Q$ will have different sizes... that's what's giving me pause. Am I missing something? I've come up with an argument that $V(PA) = V(A)$, but right multiplication is causing me an issue. Aug 4, 2016 at 16:00
• You have an intuition for the $A:R^n\rightarrow R^n$ case, so you also have an intuition for the $A:R^n\rightarrow R^{n+k}$ case by just adding an $R^k$ factor to your target space. So you have an intuition for the $n<m$ case by considering the image of $A$. For the $n>m$ case, you then have the same intuition for $A^T: R^m\rightarrow R^n$.
– Mud
Aug 4, 2016 at 16:42
• @Mud This was very helpful in easing my thinking about it, thank you kindly! Aug 4, 2016 at 19:25

Let $A$ be the matrix in question, writing $D(A) = \sqrt{\det(A^T A)}$ and $V(A)$ for the desired $n$-volume; we want to see that $D(A) = V(A)$. Note that the claim is easy if $m < n$, since the rank of $A^T A$ is bounded above by the minimum of rank $A$ and rank $A^T$, so we assume $m \ge n$.

Choose orthogonal matrices $P$ and $Q$ of sizes $m \times m$ and $n \times n$. Now $$D(PAQ) = \sqrt{\det((PAQ)^T (PAQ))} = \sqrt{\det(Q^T A^TA Q)} = \sqrt{\det(A^T A)} = D(A).$$ I claim that $V(PA) = V(A)$. Indeed, write $$A = (v_1 \; v_2 \; \dots \; v_n)$$ where the $v_i$ are columns. It is now possible to choose $v_i$ for $n+1 \le i \le m$ in such a way that $V(A) = |\det (v_1 \; v_2 \; \dots \; v_m)|$; in fact, take $v_{n+1}$ to be of unit length and orthogonal to the span of the columns of $A$, and then repeat. Now, since $P$ preserves inner products (and norms), $$V(A) = \left |\det (v_1 \; v_2 \; \dots \; v_m)\right| = \left|\det (Pv_1 \; Pv_2 \; \dots \; Pv_m)\right| = V(Pv_1 \; Pv_2 \; \dots \; Pv_n) = V(PA).$$ Now write $A = URW$ using $SVD$, where $U, W$ are orthogonal and $R$ is rectangular diagonal with non-negative entries. Then $$V(A) = V(U^T A) = V(RW) = V(R).$$ The last equality follows since $V(W) = 1$, and if the diagonal entries of $R$ are $d_1, \dots, d_n$, then the column vectors of $RW$ are $d_1$ times longer than $W$'s in the 1st dimension, $d_2$ times longer in the 2nd dimension, etc. Thus $V(A) = V(R) = D(R) = D(URW) = D(A)$, where $V(R) = D(R)$ comes down to checking the claim for $n$-cells with edges in the axes.

Similarly to the axiomatic definition of the determinant, you might check that the provided formula satisfies linear homogeneity (for positive scaling of each column, because unlike the determinant it gives us non-oriented volume) shear invariance (adding one column to another doesn't change the volume) and is normed (hypercubes have volume 1).

Most textbooks on linear algebra show that these properties suffice to determine a unique function, which then must be the (non)oriented volume, since the (non)oriented volume certainly satisfies all the above axioms

• This was very helpful in demonstrating to me how some rigorous framework can help build up and fortify intuition. It's such a shame that I can't accept multiple answers... Aug 4, 2016 at 19:27

A bit late, but I have been thinking in this ideas and I want to share my conclusions. What follows is not very formal, but I think is good for visualization. Please, let me know if there is a mistake.

In $$\mathbb{R}^m$$ we have a natural basis, the canonical one, and we have a natural volume form, the classical determinant. Indeed, we can think that the choice of this basis gives rise to an inner product or scalar product (by establishing this basis as orthonormal) whose associated volume form is the determinant. That is, the determinant measures the same volume on parallelepipeds that we would obtain by measuring the edges and angles with the inner product. The volume form is not determined, therefore, by the basis but by a class of basis with the same associated inner product (one basis can be obtained from the other by an orthogonal transformation).

But we want also to compute "intermediate volumes": lengths, areas, 3-volumes,... up to the $$m$$-volume, and this can be done by using the inner product (classical scalar product). But there is a formula to compute them directly. If $$A$$ is the matrix whose columns are the $$n$$ vectors $$v_1,\ldots,v_n \in \mathbb{R}^m$$ whose $$n$$-volume want to be computed then the $$n$$-volume is $$D(A):=\sqrt{\det(A^T A)}.$$

Observe that this formula includes as particular cases the 1-volume or "length", and the $$m$$-volume (determinant of a square matrix).

In order to prove this formula, we can consider that the $$n$$ vectors $$v_1,\ldots,v_n \in \mathbb{R}^m$$ are linearly independent (if not, their $$n$$-volume would be zero) and look for an orthogonal transformation $$T$$ of $$\mathbb{R}^m$$ (with matrix $$U$$ in the canonical basis) that sends the $$n$$-dimensional subspace spanned by them to $$span(e_1,\ldots,e_n)$$ being $$\{e_i\}$$ the canonical basis of $$\mathbb{R}^m$$. Since the transformation is orthogonal, the $$n$$-volume should be the same, that is, $$Vol(A)=Vol(UA).$$ But this volume, since is "inside" $$\mathbb{R}^n\subset \mathbb{R}^m$$ can be computed classically as $$\det(trunc(UA))$$ where $$trunc$$ means selecting the first $$n$$ rows (the others are null).

And finally, observe that $$\det(trunc(UA))=\sqrt{\det(trunc(UA)^Ttrunc(UA))}=$$ $$=\sqrt{\det((UA)^T UA)}=\sqrt{\det(A^T A)}$$ since the truncated part is a 0 block matrix and since $$U^T U=Id$$.

So $$Vol(A)=\sqrt{\det(A^T A)}$$.

Different metric.

I we had started with a different inner product $$\langle-,-\rangle$$ (or with another basis, not corresponding to an orthogonal transformation of the canonical one), I think the formula would be nearly the same: $$Vol(A)=\sqrt{\det(A^T G A)}$$ where $$G$$ is the matrix of the inner product, that is, $$G=(g_{ij})$$ with $$g_{ij}=\langle e_i, e_j \rangle$$.