Abstract Nonsense

Crushing one theorem at a time

Halmos Section 29,30 and 31: Multilinear Forms, Alternating Multilinear Forms, Alternating Multilinear Forms of Maximal Degree


Point of post: In this post we solve the problems given at the end of the sections 29,30 and 31 in Halmos’s book

1.

Problem: Give an example of a vector space \mathscr{V} and a multilinear form on \mathscr{V} which is skew-symmetric but not alternating.

Proof: By virtue of a previous theorem it’s clear that that the ground field for \mathscr{V} have characteristic 2. So, let’s just consider \mathscr{V} to be \mathbb{Z}_2 over itself. Then, we can consider

K:\mathbb{Z}_2\boxplus\mathbb{Z}_2\to\mathbb{Z}_2:(x,y)\mapsto xy

This is evidently bilinear since

K(\alpha x+\beta x',y)=(\alpha x+\beta x')y=\alpha xy+\beta xy'=\alpha K(x,y)+\beta K(x',y)

since it’s symmetric it follows that K is bilinear. Also, it’s clear that it’s skew-symmetric since (recalling that -1=1 in \mathbb{Z}_2 \pi K=\pm K=K. It’s not alternating however since K(1,1)=1\ne 0.

2.

Problem: Give an example of a non-zero alternating k-linear form K on an n-dimensional space (k<n) such that K(x_1,\cdots,x_n)=0 for some linearly independent set of vectors x_1,\cdots,x_n.

Proof:

Plan of attack: What if we considered a non-zero alternating k+1-form but fixed the last entry. So, to be explicit what if we looked at some non-zero K\in\text{Alt}_{k+1}\left(\mathscr{V}\right), which we know exists by prior theorem since k+1\leqslant n. Then, we could find some x_1^0,\cdots,x_k^0,x_0 (the superscripts are just meant so you don’t confused these fixed vectors with the variable vectors x_1,\cdots,x_k which are soon to be used) such that K\left(x_1^0,\cdots,x_k^0,x_0\right)\ne 0. So, what if we considered \tilde{K}\in\text{Alt}_{k}\left(\mathscr{V}\right) given by \tilde{K}\left(x_1,\cdots,x_k\right)=K\left(x_1,\cdots,x_k,x_0\right). Indeed this function is multilinear since K is multilinear and alternating since K is alternating. Moreover, K is non-zero since \tilde{K}\left(x_1^0,\cdots,x_k^0\right)=K\left(x_1^0,\cdots,x_k^0,x_0\right)\ne 0. But, what if we picked a set of k vectors \{x_1,\cdots,x_k\} which are linearly independent but for which \{x_1,\cdots,x_k,x_0\} isnt? Then \tilde{K}\left(x_1,\cdots,x_k\right)=K\left(x_1,\cdots,x_k,x_0\right)=0 since K is alternating and \{x_1,\cdots,x_k,x_0\} is linearly dependent. So with this in mind let’s consider an actual:

Anti-climactic Example

Consider

K:\mathbb{R}^3\boxplus\mathbb{R}^3\to\mathbb{R}:\left(\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix},\begin{bmatrix}y_1\\ y_2\\ y_3\end{bmatrix}\right)\mapsto \det\begin{bmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 &0\end{bmatrix}

clearly then K\in\text{Alt}_2\left(\mathbb{R}^3\right) and K is non-zero but

\displaystyle K\left(\begin{bmatrix}1\\ 0\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}\right)=\det\begin{bmatrix}1 & 0 &1\\ 0 & 1& 1\\ 0 & 0 & 0\end{bmatrix}=0

even though \left\{\begin{bmatrix}1\\ 0\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}\right\} is a linearly independent set of vectors.

3.

Problem: What is the dimension of all symmetric k-linear forms (k<n)? What about the skew-symmetric? What about the alternating ones?

Proof: Personally, and maybe this problem just psyched me out, but I thought this was a tough one to prove. We begin by classifying the skew-symmetric forms.

Claim:Let \mathscr{V} be an n-dimensional space, then the dimension of \displaystyle \text{SkewSym}_k\left(\mathscr{V}\right)={n\choose k}.

Proof: Fix a basis \{x_1,\cdots,x_n\}=\mathcal{B} for \mathscr{V}. Then, for each \{\ell_1,\cdots,\ell_k\} with \ell_1<\ell_2<\cdots<\ell_k let K be the unique k-linear form on \mathscr{V} such that

\displaystyle K_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)=\prod_{j=1}^{k}\delta_{\ell_j,r_j}

Then, let J_{\ell_1,\cdots,\ell_k}=\text{Asym}\left(K_{\ell_1,\cdots,\ell_k}\right) where \text{Asym} is the “antisymmetrizer function” described in the post on skew-symmetric forms. Note then that if \{\ell_1,\cdots,\ell_k\}\ne\{r_1,\cdots,r_k\} then J_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)=0 (note that this no longer evaluates to zero on tuples who have the same elements as \{x_{\ell_1},\cdots,x_{\ell_k}\} but a possibly different order of arrangement) and that K(x_{\ell_{\pi(1)}},\cdots,x_{\pi(\ell_k)})=\text{sgn}(\pi).  We claim that

\mathscr{B}\overset{\text{def.}}{=}\text{span}\left\{J_{\ell_1,\cdots,\ell_k}:\{\ell_1,\cdots,\ell_k\}\subseteq[n]\right\}=\text{SkewSym}_k\left(\mathscr{V}\right)

To see that suppose that

\displaystyle \sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdots<\ell_k}\alpha_{\ell_1,\cdots,\ell_1}J_{\ell_1,\cdots,\ell_k}=\bold{0}

Note then that for each \{r_1,\cdots,r_k\}\subseteq[n] with r_1<\cdots<r_k we see that

\displaystyle \alpha_{r_1,\cdots,r_k}=\sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdots<\ell_k}\alpha_{\ell_1,\cdots,\ell_k}K_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)=0

from where linear independence follows. Now, to see that \text{span }\mathscr{B}=\text{Alt}_k\left(\mathscr{V}\right) we prove that for any K\in\text{Alt}_k\left(\mathscr{V}\right) that

\displaystyle K=\sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdot<\ell_k}K\left(x_{\ell_1},\cdots,x_{\ell_k}\right) J_{\ell_1,\cdots,\ell_k}\quad\mathbf{(1)}

to do this it suffices to show that the right hand side \mathbf{(1)} equals the left hand side for all elements of \mathcal{B}^k. To do this we let (x_{r_1},\cdots,x_{r_k})\in\mathcal{B}^k be arbitrary. If any of the coordinates of the tuples are equal then both sides of \mathbf{(1)} are zero since both sides are skew-symmetric and thus alternating, so assume not. Then, there exists some \pi\in S_k such that r_{\pi(1)}<\cdots<r_{\pi(k)} we note then that

\displaystyle K(x_{r_1},\cdots,x_{r_1})=\text{sgn}(\pi)K(x_{r_{\pi(1)}},\cdots,x_{r_{\pi(k)}})=\sum_{\{x_{\ell_1},\cdots,x_{\ell_k}\}\subseteq\mathcal{B}:\ell_1<\cdots<\ell_k}K\left(x_{\ell_1},\cdots,x_{\ell_k}\right) J_{\ell_1,\cdots,\ell_k}\left(x_{r_1},\cdots,x_{r_k}\right)

Thus, noting that the number of \{\ell_1,\cdots,\ell_k\}\subseteq[n] for which \ell_1<\cdots<\ell_k is \displaystyle {n\choose k} finishes the argument. \blacksquare
Corollary: If the characteristic of the ground field for \mathscr{V} has characteristic greater than two then \displaystyle \dim_F\text{Alt}_k\left(\mathscr{V}\right)={n\choose k}

Next we prove the result for symmetric forms, which for some reason I found considerably easier

Claim If \mathscr{V} is an n-dimensional space then \displaystyle \dim_F\text{Sym}_k\left(\mathscr{V}\right)={n+k-1\choose k}

Proof: Fix a basis \{x_1,\cdots,x_n\}=\mathcal{B}. It’s clear upon expansion of an arbitrary k-linear form though that each such form is really just the linear combination of monomials of the form \alpha_{1,f(1)}\cdots\alpha_{k,f(k)} and that these monomials are linearly independent symmetric forms. The claim follows by noticing that the number of such monomials is \displaystyle {n+k-1\choose k}. The conclusion follows. \blacksquare

References:

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

Advertisements

November 18, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: