Use of determinants for vectors - what is the intuition behind it?

Question

I am currently taking a Calculus 3 class. We just began using determinants in the study of vectors. I have some questions as to the apparent 'arbitrariness' of how we use determinants in the study of vectors. How does the determinant of three vectors give the volume of the parallelepiped they form? Similarly, how does the determinant of two vectors yield their cross product?

The professor said that that is the way they have been defined. But such a definition did not seem self-explanatory, and I figured there must have been a deeper reason, perhaps owing to some physical intuition. Also, if determinants had been defined in such a precise manner so that our answer would match what we get using other methods (such as |A||B|sinθ), then how is that definition valid for other more abstract uses of determinants?

I have not yet taken Linear Algebra, so perhaps the reason will become clear to me once I do, but I thought I should know before I proceeded. Thanks.

Answer 1

My answer will be http://mathinsight.org/relationship_determinants_area_volume

To put short, determinants relates to the volume because determinants IS a combination of cross product and dot product (by definition of cross product), and the two product have a geometric meaning to volume and area. However, I will also provide a non-essential, but helpful, supplement below to make it even (hopefully) more intuitive.

The more detailed and picture rich explanation can be found in the link above. However, I have to make a supplement definition of cross product here, as most people only get an unintuitive and shitty definition that only works for $\mathbb R^3$.

Let $u_1,...,u_{n-1} \in \mathbb R^n$, we define $$\times(u_1,...,u_{n-1}) := w$$ to be the unique vector $w$ such that

$$\det[u_1 | u_2 | ... | u_{n-1} | x] = w \cdot x$$

for all $x \in \mathbb R^n$. Specifically, for $n= 3$, the handy dandy formula to calculate the cross product, is not just a formula, it is THE definition. I will give a motivation below.

For simplicity, let $n= 3$, as it is exactly the same for higher dimensions, you just need to write more. Cross product aims at creating a vector that is perpendicular to two give vectors $u_1, u_2$. Assuming you understand the canonical inner product aka dot product $u \cdot v$, we want to find a vector $w$ where $$w \cdot u_1=0 = w\cdot u_2.$$

In fact any $x = a_1 u_1+a_2 u_2$ should have $w \cdot x = 0$. We now have a function

$$f(x) = w \cdot x$$

The next little bit requires a bit of linear algebra. We already know a function $g(x)$ such that $g(x) = 0 \iff x = a_1 u_1+a_2 u_2$, and that function is the determinant function

$$g(x) = \det M(x)$$

where $M(x) = [u_1 | u_2 | x]$. That is the $3 \times 3$ matrix where the first column is $u_1$, second is $u_2$, and the third is our input vector $x$. Or with $u_k = (u_{k1}, u_{k2}, u_{k3})$ and $x = (x_1,x_2,x_3)$:

$$M(x) = \begin{bmatrix} u_{11} & u_{21} & x_1\\u_{12} & u_{22} & x_2\\u_{13} & u_{23} & x_3\\ \end{bmatrix}$$

So we have $$f(x) =w \cdot x = 0 \iff g(x) = \det \begin{bmatrix} u_{11} & u_{21} & x_1\\u_{12} & u_{22} & x_2\\u_{13} & u_{23} & x_3\\ \end{bmatrix} = 0$$

So why don't we just let $f(x) = g(x)$, that is take the vector $w$ where $w \cdot x = \det M(x)$ at all values rather than at the roots only. That's how we will define cross product.

It turns out this definition leads to many properties we love about cross product, and it extends to all dimensions, providing a tool to make perpendicular vectors.

Continue in $n= 3$. Again, $\det M(x) = (u_1 \times u_2) \cdot x$ for all $x \in \mathbb R^3$. It turns out the volume of the parallelopiped spanned by $a,b,c \in \mathbb R^3$, up to a sign, is

$$(a\times b) \cdot c$$

which is $\det[a|b|c]$, where $[a|b|c]$ is the $3 \times 3$ matrix with $a,b,c$ as columns, by definition. $(a \times b)$ contributes the bottom area, and $c$ contributes to the height. Explained in the link above.

You might have rows in your definition, but that's ok, as $\det M = \det M^T$. That is if you switch the rows with the columns, the determinant doesn't change.

P.S. Determinants is defined as the unique map that reverses signs when you switch two rows/columns, is multilinear, and preserves common factors in each row/column. The property I described about determinant being 0 is one possible motivation of why we defined cross product that way.

Answer 2

Cross products are extremely unintuitive because they have simply been defined to work for our purposes, as your professor said. We basically made up $i$, $j$, and $k$. Then we said

$$i\times j=k,\: j\times i=-k$$ $$j\times k=i,\: k\times j=-i$$ $$k\times i=j,\: i\times k=-j$$

as well as $i\times i=j\times j=k\times k=0$. So we get the orthogonality relations we are so familiar with for cross products. Then, for $a=a_1i+a_2j+a_3k$ and $b=b_1i+b_2j+b_3k$, we get

$$a\times b=a_1i\times(b_1i+b_2j+b_3k)+a_2j\times(b_1i+b_2j+b_3k)+a_3k\times(b_1i+b_2j+b_3k)$$

$$=a_1b_1(i\times i)+a_1b_2(i\times j)+a_1b_3(i\times k)+a_2b_1(j\times i)+a_2b_2(j\times j)+a_2b_3(j\times k)+a_3b_1(k\times i)+a_3b_2(k\times j)+a_3b_3(k\times k)$$

$$=a_1b_2k-a_1b_3j-a_2b_1k+a_2b_3i+a_3b_1j-a_3b_2i$$

$$=(a_2b_3-a_3b_2)i-(a_1b_3-a_3b_1)j+(a_1b_2-a_2b_1)k$$

$$=\det\pmatrix{i&j&k\\a_1&a_2&a_3\\b_1&b_2&b_3}$$

Perhaps the determinant was purposely defined with this in mind, I honestly don't know enough about that or which came first.