# Abstract Nonsense

## Characterization of Alternating Multilinear Forms, and the Determinant

Point of post: In this post we give a characterization of alternating forms in terms of the values they take on any basis, and with this we characterize the determinant as the unique alternating multilinear form $K$ on the columns of a matrix $A$ such that $K\left(I\right)=1$ where $I$ is the identity matrix.

Motivation

So, in the last few posts we’ve gone from discussing $n$-linear forms on $\mathscr{V}_1\oplus\cdots\oplus \mathscr{V}_1$, to discussing $n$-forms on just $\mathscr{V}^n$ so that we could discuss symmetric and skew-symmetric forms, to finally discussing alternating forms, all with the motivation to somehow “deal” with the determinant function. So, now that we’re done with these discussion it would seem criminal to not devote an entire thread to the repercussions have on the function $\det$. Thus, in this post we give a theorem which basically says that a skew-symmetric form is determined entirely by one value, (it’s value on some basis) and from this prove the oft stated “alternate definition” of the determinant function as the “unique function $f$ of the columns of a matrix which is linear in each column, skew-symmetric, and has $f\left(I\right)=1$.

Characterization of $n$-forms

Now that we have dealt enough with alternating $n$-forms to get a “feel” for them, we can safely state some obvious theorems, which trim away the fluff in the evaluation of such forms. For example, we noticed in the last post that when one expands an alternating $n$-form via some basis that “a lot” of the terms disappear via the alternating quality of the form. We now formalize how may “a lot” is

Theorem: Let $\mathscr{V}$ be a $m$-dimensional $F$ space with basis $\{x_1,\cdots,x_m\}$ then

$\displaystyle K\left(\sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right)=\sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}K(x_1,\cdots,x_m)$

Proof: Recall that by definition

$\displaystyle K\left(\sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right)=\sum_{j_1=1}^{m}\cdots\sum_{j_m}^{m}\prod_{k=1}^{m}\alpha_{k,j_k}K\left(x_{j_1},\cdots,x_{j_m}\right)\quad \mathbf{(1)}$

A little thought shows that the right hand side $\mathbf{(1)}$ can be rewritten as

$\displaystyle \sum_{f\in[m]^{[m]}}\prod_{k=1}^{m}\alpha_{k,f(k)}K\left(x_{f(1)},\cdots,x_{f(m)}\right)\quad\mathbf{(2)}$

where $[m]^{[m]}=\left\{f\mid f:[m]\to[m]\right\}$. But, clearly we may partition $[m]^{[m]}$ into two blocks, namely

$B_1=\left\{f\in[m]^{[m]}:f\text{ is an injection}\right\}$

and

$B_2=\left\{f\in [m]^{[m]}:f\text{ is not an injection}\right\}$

recall though that $f\in[m]^{[m]}$ is an injection if and only if it’s a bijection so that $B_1=S_m$ and thus $B_2=[m]-S_m$. Thus, it follows that $\mathbf{(2)}$ may be rewritten as

$\displaystyle \sum_{f\notin S_m}\prod_{k=1}^{m}\alpha_{k,f(k)}K\left(x_{f(1)},\cdots,x_{f(m)}\right)+\sum_{\pi\in S_m}\prod_{k=1}^{m}\alpha_{k,\pi(k)}K\left(x_{\pi(1)},\cdots,x_{\pi(m)}\right)\quad\mathbf{(3)}$

but notice though that by definition of an alternating $n$-form that if $f\in[m]^{[m]}$ is not an injection then $K\left(x_{f(1)},\cdots,x_{f(m)}\right)=0$ so that the first term in $\mathbf{(3)}$ vanishes. But, recalling that for $\pi\in S_m$ we have that

\begin{aligned}K\left(x_{\pi(1)},\cdots,x_{\pi(m)}\right) &=\left(\pi^{-1} K\right)\left(x_1,\cdots,x_m\right)\\ &=\text{sgn}(\pi^{-1})K\left(x_1,\cdots,x_m\right)\\ &=\text{sgn}\left(\pi\right) K\left(x_1,\cdots,x_m\right)\end{aligned}

and thus with both of these facts in mind we see that $\mathbf{(3)}$ may be rewritten as

$\displaystyle \sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}K\left(x_1,\cdots,x_m\right)$

as desired. $\blacksquare$

Theorem: Let $\mathscr{V}$ be a $m$-dimensional $F$-space with basis $\{x_1,\cdots,x_m\}$, then for any $\alpha_0\in F$ there is a unique alternating $n$-form $K$ for which $K(x_1,\cdots,x_m)=\alpha_0$.

Proof: We claim that the function

$\displaystyle K\left(\sum_{j_1=1}^{m}\alpha_{1,j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right)=\sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}\alpha_0$

is such a function as described in the theorem. It’s evidently multilinear and $K(x_1,\cdots,x_m)=\alpha_0$, so it suffices to prove that it’s alternating. To do this we note that for any transposition $\tau\in S_m$ we have that $S_m=A_m\cup \tau A_m$ where, as always, $\tau A_m=\left\{\tau\pi:\pi\in A_m\right\}$. Thus, suppose that

$(x_1,\cdots,\underbrace{x}_{k^{\text{th}}},\cdots,\underbrace{x}_{\ell^{\text{th}}},\cdots,x_m)\in\mathscr{V}^m$

then we can see using our previous observation with $\tau=(k,\ell)$ to see that

$\displaystyle \sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}\alpha_0=\alpha_0 \sum_{\pi\in A_m}\left(\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}+\text{sgn}(\pi\tau)\prod_{k=1}^{m}\alpha_{\tau(k),\pi(k)}\right)\quad\mathbf{(1)}$

But, noting that $\text{sgn}(\pi)=1$ and $\text{sgn}(\pi\tau)=-1$ for all $\pi\in A_m$ and making the observation that since $\tau$ only switches the two slots which are identical (and in particular the coefficients of $x_1,\cdots,x_m$ are identical) we see that

$\displaystyle \prod_{k=1}^{m}\alpha_{\tau(k),\pi(k)}=\prod_{k=1}^{m}\alpha_{k,\tau(k)}$

we may conclude that $\mathbf{(1)}$ is equal to

$\displaystyle \alpha_0\sum_{\pi\in A_m}\left(\prod_{k=1}^{m}\alpha_{k,\pi(k)}-\prod_{k=1}^{m}\alpha_{k,\pi(k)}\right)=0$

and since $x\in\mathscr{V}$ and $k,\ell\in[m]$ were arbitrary the conclusion follows.

Now, uniqueness is clear by the previous theorem. $\blacksquare$

With this theorem under our belts and the last two theorems of the last post we may confidently state that:

Corollary: The function

$\displaystyle \det\left(\begin{bmatrix}\alpha_{1,1}\\ \alpha_{1,2}\\ \vdots\\ \alpha_{1,m-1}\\ \alpha_{1,m}\end{bmatrix},\cdots,\begin{bmatrix}\alpha_{m,1}\\ \alpha_{m,2}\\ \vdots\\ \alpha_{m,m-1}\\ \alpha_{m,m}\end{bmatrix}\right)=\sum_{\pi\in S_n}\text{sgn}\pi\prod_{k=1}^{m}\alpha_{k,\pi(k)}$

is the unique function alternating $m$-linear form $K$ on $F^m$ (where $F$ is some field) for which

$\displaystyle K\left(\begin{bmatrix}1_F\\ 0\\ \vdots\\ 0\\ 0\end{bmatrix},\cdots,\begin{bmatrix}0\\ 0\\ \vdots\\ 0\\ 1_F\end{bmatrix}\right)=1_F$

And thus, the determinant is the unique alternating $m$-linear forms on the columns of the elements of $\text{Mat}_m\left(F\right)$ for which $\det(I)=1_F$. Moreover,

$\displaystyle \det\left(\begin{bmatrix}\alpha_{1,1}\\ \alpha_{1,2}\\ \vdots\\ \alpha_{1,m-1}\\ \alpha_{1,m}\end{bmatrix},\cdots,\begin{bmatrix}\alpha_{m,1}\\ \alpha_{m,2}\\ \vdots\\ \alpha_{m,m-1}\\ \alpha_{m,m}\end{bmatrix}\right)=0$

if and only if $\left\{\begin{bmatrix}\alpha_{1,1}\\ \alpha_{1,2}\\ \vdots\\ \alpha_{1,m-1}\\ \alpha_{1,m}\end{bmatrix},\cdots,\begin{bmatrix}\alpha_{m,1}\\ \alpha_{m,2}\\ \vdots\\ \alpha_{m,m-1}\\ \alpha_{m,m}\end{bmatrix}\right\}$ forms a linearly dependent set. $\blacksquare$

References:

1. 1. Halmos, Paul R. ” Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

November 15, 2010 -

## 3 Comments »

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2. […] By virtue of a previous theorem it’s clear that that the ground field for have characteristic . So, let’s just […]

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3. […] Now that we have the definition of the derivative for mappings it’s time to get our hands a little dirty and compute something. In particular we aim at proving that the wide sweeping class of multilinear function on spaces of the form are everywhere differentiable and compute their derivative. From this we will be able to recover as a corollary a lot of particulary (and important) functions are differentiable, in particular linear trnasofrmations, the functions of the form given by and , the usual inner product on , and the determinant. […]

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