Abstract Nonsense

Crushing one theorem at a time

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

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November 15, 2010 - Posted by | Algebra, Halmos, Linear Algebra | , ,

3 Comments »

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