Intutive meaning of $\det(AB)=\det(A) \det(B)$.

Question

If we take determinant as volume of unit cube let say A than $\det(A)=1$ as its volume is 1. Now let take another unit cube B and if we put both cubes side by side than then $\det(A) \det(B)=1*1=1$ only. So what is the physical meaning of the $\det(AB)$ in the above example such that $\det(AB)=1$. Means how to put the cubes such that its $\det$ becomes 1.

Can any one help me to understand this may be I am missing some point?

Answer 1

Start with a unit cube, volume 1.

Multiply by matrix $A$. This will stretch, rotate and warp the cube, so it now has volume equal to $\det A$. Air inside the cube is now that much less dense.

Now sit a fresh unit cube next to it, and multiply both by $B$. Air inside the new cube is now stretched to a volume $\det B$. With the same $B$, the first cube, which did have volume $\det A$, now has volume $\det B$ times as much, so it is $\det A\det B$.

On the other hand, the first cube has gone through $A$ then $B$, so it has gone through $BA$, and its volume must be $\det BA$.

Answer 2

Let $X$ be a unit cube in $\mathbb{R}^n$. If $A:\mathbb{R}^n\to\mathbb{R}^n$ is a linear transformation, then $\det{A}$ is the (signed) volume of the parallelotope $A(X)$. In other words, $\det A$ is the factor by which volume in $\mathbb{R}^n$ is scaled after we apply $A$. So if we have another linear transformation $B$, and we want to know $\det(A\circ B)$, with a little bit of thought we can just multiply the two scaling factors together.

Answer 3

Well, I'm not sure this gives intuition, but you could ask yourself what properties you want from a function that computes signed volume and go from there.

Suppose we have such a function $d: \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}$. That is, given points $x_1,...,x_n$, the volume of the parallelepiped whose edges are the lines between $0$ and $x_k$ is given by $d(x_1,...,x_n)$.

We expect that it would be linear in each variable, that is $d(x_1,...,x_k+y_k,...,x_n) = d(x_1,...,x_k,...,x_n) + d(x_1,...,y_k,...,x_n)$ and $d(x_1,...,\lambda x_k,...,x_n) = \lambda d(x_1,...,x_k,...,x_n)$. In other words, $d$ is multilinear.

We expect that switching the order of two variables switches the sign of $d$, that is $d(\cdots, x_i,\cdots, x_j, \cdots) = -d(\cdots, x_j,\cdots, x_i, \cdots)$. In other words, $d$ is alternating.

If you grind through the details, you can show that you must have $d(x_1,...,x_n) = \det \begin{bmatrix} x_1 & \cdots & x_n\end{bmatrix} d(e_1,...,e_n)$, where $\det \begin{bmatrix} x_1 & \cdots & x_n\end{bmatrix}$ is the usual formula involving permutations and signs.

The point is that by just requiring multilinearity and alternation, we must have a scaled version of the usual determinant formula. This may not seem like a big deal, but it means the notion of volume is encapsulated by being multilinear and alternating (and some reference volume).

At the risk of mind-numbing repetition, if $\delta$ is another alternating, multilinear function (on $n$ copies of $\mathbb{R}^n$), we must have $\delta(x_1,...,x_n) = \alpha \det \begin{bmatrix} x_1 & \cdots & x_n\end{bmatrix} $, where $\alpha = \delta(e_1,...,e_n)$. To hammer it home, the space of such functionals has dimension one.

We expect that the 'unit cube' has volume one, so we single out one of these objects and call it $\operatorname{vol}$. The above considerations shows that $\operatorname{vol}(x_1,...,x_n) = \det \begin{bmatrix} x_1 & \cdots & x_n\end{bmatrix} $.

If we have a matrix $B$, then the volume of the parallelepiped whose edges are $B e_1,...,B e_n$ is given by $\operatorname{vol}(B e_1,...,B e_n) = \det B$.

Now consider another such object, $\lambda_A(x_1,...,x_n) = \operatorname{vol}(Ax_1,...,Ax_n)$. We can check that it is alternating and multilinear, so we must have $\lambda_A = \alpha \operatorname{vol}$ for some $\alpha$. We can figure out $\alpha$ by looking at the 'unit cube', that is, $\lambda_A(e_1,...,e_n) = \operatorname{vol}(Ae_1,...,Ae_n) = \det A = \alpha \operatorname{vol}(e_1,...,e_n) = \alpha$. In other words, $\lambda_A = (\det A) \operatorname{vol}$.

In particular, we have shown $\det (AB) = \lambda_A(B e_1,...,B e_n) = \det A \operatorname{vol}(B e_1,...,B e_n) = \det A \det B$.

In other words, instead of thinking of $\det (AB)$, think of the scaled volume operator $\lambda_A$ operating on $B$ (or rather the columns of $B$), and that $\lambda_A$ is essentially $\det$ except for a scaling factor.