In my matrix analysis course, we are seeking to understand the idea behind determinants. In class, my professor mentioned that
"The determinant of an $n \times n$ matrix can be thought of as an alternating $n$-linear function of its column vectors."
For clarity, an $n$-linear form is alternating if $x_i=x_j \Rightarrow f(x_1, \ldots, x_n) =0$ for $i\neq j$.
This idea is one that I can't quite wrap my head around. I understand that the determinant can be thought of as a scaling factor for the volume generated by basis vectors, but beyond that, I'm struggling to see how determinants relate to multilinear maps. I am also stuck on why the alternating condition is important.
I found this question on MSE, but it only confused me more.
This question provided a little more insight, but I feel like I still don't have all the prerequisite knowledge to effectively understand everything.
Any help would be appreciated.
Best Answer
Let $V=K^n$ be the vector space of dimension $n$. We can write a $n\times n$ matrix in the following way: $$ M=\left(\begin{array}{cccc} \mid&\mid&&\mid\\ v_1&v_2&\cdots&v_n\\ \mid&\mid&&\mid \end{array}\right), $$ and see each column as a vector in a $n$-dimensional vector space. Thus let $d$ be a multilinear map: $$ d:\underbrace{V\times\cdots\times V}_\text{$n$ times}\to K, $$ then we can compute $d(v_1,v_2,\ldots,v_n)$, using the entries of $M$. If you write $d$ as the determinant function, can you see that the defined map will indeed be multilinear and alternating? (this justifies the affirmation of your professor.)
Edit: Better writing: $$ \det:(v_1,\ldots,v_n)\in V\times\cdots\times V\mapsto\det\left(\begin{array}{cccc} \mid&\mid&&\mid\\ v_1&v_2&\cdots&v_n\\ \mid&\mid&&\mid \end{array}\right)\in K. $$
Extra: How many maps $f:\underbrace{V\times\cdots\times V}_\text{$n$ times}\to K$ that are multilinear and alternating can exist, if $V$ has dimension $n$?