How to see from definition that determinant is just volume?

Question

The determinant can be defined recursively by Laplace expansion (LE), or equivalently, directly defined by Lebniz formula (LF) using parity of permutation.

My question is:

How do we see that these calculations indeed carry out the volumn of the parallelopiped expanded by row vectors of the determinant?

I found some explanations, all of which are not satisfactory for me.

• This answer utilizes diagnolization of matrix and then we see that the product of the eigenvalues is indeed the volume.
• This interactive textbook first defines the determinant to be an abstract multilinear function, and then work out the calculation formula, and finally proves by: The only function satisfying the defining properties of determinant is the determinant itself, and the volume of vectors satisfies these properties.

These explanations, without exception, all circumvents the calculation formulas (LE, LF). But I'm wondering this. Determinant is first discovered when solving system of linear equations and defined by LF. How do the earliest researchers, without knowledges of diagnalization, abstract multilinear functions, etc., sees that this definition (LE or LF) is just the volume of vectors?

One more thing. Please do not use the definition of cross product of vectors. Because this operation is also best calculated by formal determinant (as I see it). In fact, this question first came to me when I was thinking: Why the cross product, or the vector orthogonal to two given vectors with length equal to the area of the parallelogram, can be calculated by $$ \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix}$$

Answer 1

How do the earliest researchers, without knowledges of diagnalization, abstract multilinear functions, etc., sees that this definition (LE or LF) is just the volume of vectors?

Here's a slide I made that gives a simple explanation for the determinant of a 2 by 2 matrix.

A similar but more laborious argument could be made for the determinant of a 3 by 3 matrix. At that point, we can guess that this interpretation holds for the determinant of an $n \times n$ matrix

Answer 2

TLDR; You can use row operations on a matrix to convert it to a diagonal matrix, without changing the determinant. Geometrically, this corresponds to shear operations on a parallelepiped, transforming it to cuboid with sides parallel to axes. In the end, the volume of a cuboid is the product of side lengths and the determinant of a diagonal matrix is the product of diagonal values.

Let's assume that we agree with the following statement about $n$-dimensional volume of parallelepiped on vectors $x_1,\ldots,x_n$. If $x_1 = h+e$, where for $h\perp x_i$, $i=2\ldots n$ and $h\perp e$, then $$V(x_1,\ldots,x_n) = |h|\cdot V(x_2,\ldots,x_n).$$ In other words, the volume is the product of height and base area.

On the other hand, that means that if you add to $x_1$ any linear combination of $x_2,\ldots,x_n$, the volume won't change, since the volume doesn't depend on the vector $e$.

In matrix form it corresponds to the row elimination process: $$\det M = \det\begin{pmatrix}x_1\\x_2\\x_3\\\vdots\\x_n\end{pmatrix} = \det\begin{pmatrix}x_1\\x_2+\alpha_2x_1\\x_3\\\vdots\\x_n\end{pmatrix} =\det\begin{pmatrix}x_1\\x_2+\alpha_2x_1\\x_3+\alpha_3x_1\\\vdots\\x_n\end{pmatrix} =\ldots=\det\begin{pmatrix}x_1\\x_2+\alpha_2x_1\\x_3+\alpha_3x_1\\\vdots\\x_n+\alpha_nx_1\end{pmatrix} = \det M'. \!\,\!\, $$

Assume, $x_{11}\neq 0$, then you can choose $\alpha_2,\ldots,\alpha_n$, so $x'_{21}=x'_{31}=\ldots=x'_{n1}=0$: $$ M' = \begin{pmatrix} x_{11}&x_{12}&\ldots&x_{1n}\\ 0 &x'_{22}&\ldots&x'_{2n}\\ 0 &x'_{32}&\ldots&x'_{3n}\\ \ldots&\ldots&\ldots&\ldots\\ 0 &x'_{n2}&\ldots&x'_{nn} \end{pmatrix} $$

Now it's clear that on the one hand: $$ \det M' = x_{11}\det \begin{pmatrix} x'_{22}&\ldots&x'_{2n}\\ x'_{32}&\ldots&x'_{3n}\\ \ldots&\ldots&\ldots\\ x'_{n2}&\ldots&x'_{nn} \end{pmatrix} $$ On the other hand, in parallelepiped $M'$, $|h|=|x_{11}|$, so the volume is too $V(x_1,x'_2,\ldots,x'_n)=|x_{11}|V(x'_2,\ldots,x'_n)$. The formulas are the same up to sign. So both for determinant and for volume, the step of induction is the same. And since the base of the induction is also the same (|\det(a)| = |a|), we can conclude that determinant is a signed volume. “Signed” part needs more rigour from our side, but I hope you can follow how volume appears.

Answer 3

Here, we need to start from the definition of determinant.
I will show every thing in three basic definitions as below.
Definition 1: det(I)=1
Here are few things that you need to accept :

1. We know in the three dimensional space which we live, we define the volumn of the things as the volumn. And I want to expand the idea here in all dimensional space other than the nullspace.
   For example, in $R^1$ the volumn of a thing is just the length of it, and the volumn of the thing in $R^2$ is just the area of a plate, for high dimensional space we can not see, but we can imagine it.
   
2. The volumn itself has a sign, otherwise you can understand it as the value of the volumn can take negative numbers. I will expain it by a painting, suppose there is a painting in you hand, and if you flip it over, by the definition of the volumn in real life, the volumn of the painting which is just the area of the painting did not change. But actually when we look at it, we find it is different, so here i define that the volumn of the box a=can be negative.
   
3. Here it comes the fundamental idea, we define det(I) = 1, which is in the real life just equals to the volumn of the box, which this linear transformation represent. For example, the determinant of the identity matrix in $R^1$ is the volumn of the unit vector, also is just the length of it, so it is really 1; The determinant of the identity matrix in $R^2$ is just the volumn of the box of the space two orthogonal vector forms and which is just the area in two dimensional space; For a higher dimensional space $R^n$ we find the determinant is just the volumn of the box is just the volumn of the space formed by the n orthogonal unit vectors.
   

I think you know have linked the determinant of the identity matrix to the volumn of the box, such like the length of the line in $R^1$, the area in $R^2$. And I hope you have also accept the idea that the volumn of a box can be negative, because some characteristic of the volumn here is orientation determined, which is very important in the next definition.
Definition 2: det($\left[\begin{matrix}R_1^T\\R_2^T\\...\end{matrix}\right]$)=-det($\left[\begin{matrix}R_2^T\\R_1^T\\...\end{matrix}\right]$)
You can understand this by flipping the picture as I mentioned before, then you cna also link this kind of change in the determinant with the change in the volumn of the box.
Definition 3-1:det($\left[\begin{matrix}cR_1^T\\R_2^T\\...\end{matrix}\right]$)=cdet($\left[\begin{matrix}R_1^T\\R_2^T\\...\end{matrix}\right]$)
This is called the first part of the multi-linearity.
Now here are a few steps to understand this definition:

1. We know that every linear transformation can be expressed as vectors in the coordinate system, and we have known that we can deal with very standard form situation which is all the vectors are unit vectors, and they are orthogonal, in this kind of situation, we will know the determinant t=of the matrix or teh volumn of the box can both be -1 or 1, depends on the sequence of the vector.
   
2. However, in real applications, we will meet numerous different unstandard shape, this time, we need to use some rules to do the change of the shape when now we have to use the characteristic of multi-linearity.
   
3. We now have the first part of linearity, which is stretching the vector with out the change in the direction. We have to make sure in this kind of transformation, the change in the determinant keep the same pace with the volumn of the box.
   
4. Let's first find how the determinant changed in this process, we got the determinant was timed by c.
   
5. Let's then find how the volumn of the box was changed during this process, we found it is just multiply one of the vector by c, while do not change the direction of it, and also it will not change the sequence of the vector, so we know according to our definition, teh volumn of the box will be timed by c.
   
6. Now we got the change in the determinant and the change in the volumn of the box keeps the same pace.
   

Now we can make a short conculsion here, we now know that if we now have a linear transformation represented by a rectangular box, we can usse the first part of the multi-linearity to stretch the vector and turn it into a cubic like object.
But we not every matrix vector expression has orthogonal direction, so we need another powerful tool to change none orthogonal object into orthogonal ones, then it is just what the second part of the multi-linearity.
Definition 3-2:det($\left[\begin{matrix}R_1^T+R_2^T\\R_2^T\\...\end{matrix}\right]$)=det($\left[\begin{matrix}R_1^T\\R_2^T\\...\end{matrix}\right]$)
This is just the shape changing of the volumn of the box, we can also use the same way to find that the change in the determinant is zero, and if you just take the translation of one surface of the box, the volumn of the box will never change.

Final Conclusion

We prove that changing the shape of the box has the same pace of changing the determinant,through this method, we can change every symmetric shape of mbox that a matrix represent to the standard shape that the identity matrix represents.
To conclude, we got the determinant is just equal to the volumn of the box.
Hope it will help.