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.

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    $\begingroup$ An $n$-form is alternating iff the sign alternates when you swap two parameters. $\endgroup$ – copper.hat Mar 6 '18 at 4:26
  • $\begingroup$ Thank you! I was wondering why it had that name. I'm assuming that that definition is equivalent to the one I provided? $\endgroup$ – BSplitter Mar 6 '18 at 4:30
  • $\begingroup$ It is, and it is straightforward to show using linearity. Think of swapping parameters as changing the orientation, hence the parity change. $\endgroup$ – copper.hat Mar 6 '18 at 4:30
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    $\begingroup$ It's not a linear map. That would say $\det (A+B) = \det (A) + \det (B).$ It multilinear, meaning it's linear in each variable separately. For example, if you multiply one column by a constant, you multiply the determinant by that constant. When your professor said it's n-linear, that's short for saying it's multilinear in variables. $\endgroup$ – saulspatz Mar 6 '18 at 4:33
  • $\begingroup$ Thank you! I think that helps clear the idea. Let me make sure I understand: I should think of the determinant as a multilinear function that takes in column vectors and spits out a scalar. Let $c_1, c_2, \ldots, c_n$ be the column vectors of an $n\times n$ matrix $A$. Then if $\det(c_1, c_2,\ldots, c_n) = d$, then $\det(\lambda c_1, c_2, \ldots, c_n) = \lambda d$ and $\det(c_1 + v, c_2, \ldots, c_n) = \det(c_1, c_2, \ldots, c_n) + \det(v,c_2, \ldots, c_n) = d + \det(v, c_2, \ldots, c_n). $ Does this idea seem right (where $v$ is a vector with $n$ components)? $\endgroup$ – BSplitter Mar 6 '18 at 4:45

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 $n$?

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    $\begingroup$ Thank you! This helps a ton! Is there any particular reason you put quotation marks around the equals sign? Are those evaluations not truly equal? $\endgroup$ – BSplitter Mar 6 '18 at 4:53
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    $\begingroup$ @BlakeSplitter you are welcome! =) The quotation marks are just because, formally, makes no sense $d(A)$, since $d$ does not have matrices as domain. $\endgroup$ – FYY Mar 6 '18 at 4:58
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    $\begingroup$ Also, would this concept work for row vectors as well? $\endgroup$ – BSplitter Mar 6 '18 at 5:09
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    $\begingroup$ @BlakeSplitter Yes, it works! $\endgroup$ – FYY Mar 6 '18 at 5:09
  • $\begingroup$ @BlakeSplitter consider any $n$-linear and alternating function $f$. Write $f(v_1,\ldots,v_n)$ on the columns. Can you simplify the expression and write the value of $f(v_1,\ldots,v_n)$ in terms of evaluation in the canonical basis $e_1,\ldots,e_n$? (here, we need to make clear what does it means "canonical basis, and writing the column in such basis"). Now, using the same matrix and the same $f$, write in terms of the rows, let us say, $f(w_1,\ldots,w_n)$. Simplify again to the same canonical basis. Using just alternance and $n$-linearity, you will see that both expressions coincides. $\endgroup$ – FYY Mar 6 '18 at 5:18

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