# Explaining why the determinant measures the change of volume under a matrix

I would like to explain to my students that the determinant of a matrix $A$ is in fact the measure of how the volume changes under said matrix. To do that, I would like to start from the $n$-dim. cube (represented by the standard basis $e_1,\ldots,e_n$ of $\mathbb{R}^n$) and derive the formula for the volume of the parallelepiped determined by $Ae_1,\ldots,Ae_n$. Of course, I could go to the exterior algebra $\Lambda^n(\mathbb{R}^n)$ and work there, but I don't think this would be very pedagogical...

Thus, my question is: is there a down-to-earth way to recover the determinant of $A$ as the volume of the parallelepiped defined above? And if not, how can one make a compelling argument of why taking the exterior product $Ae_1\wedge\ldots\wedge Ae_n$ gives the volume?

• How do you define the volume of the parallelepiped? – mathcounterexamples.net Nov 28 '16 at 21:03
• How did you define the determinant of a matrix? One can define $\det(A)$ as the (signed) volume of the box built on the columns of $A$, normalized by $\det(I_n) = 1$. Linearity in each component is then natural to impose, and skew-symmetry (alternating condition) follows from the fact that if two sides are equal, then the volume should be 0. There is only one such function, and that is the determinant. – Catalin Zara Nov 28 '16 at 21:06
• @CatalinZara Yes, I know. They know the usual formula with signs of permutations etc, and the rules to compute it. I would like to compute the volume of the parallelepiped and find the formula they already know how to compute. – Daniel Robert-Nicoud Nov 28 '16 at 21:11
• @DanielRobert-Nicoud Have your students ever asked you why someone would come with such a convoluted definition? – Catalin Zara Nov 28 '16 at 21:13
• @mathcounterexamples.net I was afraid that question would come. It all boils down to how you define the volume, of course. Personally, I would define it using the exterior algebra, but again, I don't think it would be helpful. So the idea would probably be to start from the standard cube, give it volume $1$, and then go by multilinearity on each "argument" (the vectors defining the sides), and antisymmetry under the switches. So, basically reformulate the exterior product of $n$ vectors in elementary terms... Do you know of a better approach? – Daniel Robert-Nicoud Nov 28 '16 at 21:15

My preferred pedagogical approach is to start with examples in $2$ and $3$ dimensions. In $2D$ we can easily see that the columns of the matrix $A$ are the vectors $\vec u= A\vec i$ and $\vec v=A \vec j$ ( where $\{\vec i, \vec j \}$ is the standard basis) and it is easy to show that the ''signed'' area of the parallelogram with sides $\vec u$ and $\vec v$ is the cross product $\vec u \times \vec v = \det A$.
In a similar way we can show that the mixed product of the vectors $\vec u=A\vec i$,$\vec v=A \vec j$ and $\vec w=A\vec k$ is the volume of the of the transformed parallelepiped,and these vectors are the columns of the matrix $A$ so that $\vec u\cdot(\vec v \times \vec w)=\det A$.
Now the generalization to $n-$dimensional space can be done noting the properties of the determinant, especially the multi-linearity, and noting that we can go from the volume of a $n-$cube to an $n-$polyhedron by linear tranformations of the sides.
In 2-dimension, if you slide the side $AB$ for a parallelogram $ABCD$ along the line containing $A$ and $B$, the area remains the same. To compute the area of parallelogram $ABCD$, one can apply a shear operation to $ABCD$ and turn it into a rectangle first.
Using this sort of operation, one can show the signed-area of $ABCD$, as a function of vectors $\vec{AB}$ and $\vec{AC}$, is anti-symmetric and multi-linear. Now invite the student to express all these into linear algebra, the formula for determinant for $2 \times 2$ matrix should come up naturally.