Dimension of vector space of multilinear forms $V^n \rightarrow K$

Question

A multilinear form is a mapping

\begin{align} \Delta: V^n \rightarrow K \end{align}

where $V$ is a finite-dimensional vector space over field $K$.

It must meet the following requirements:

• First:

\begin{align} &\Delta\left(a_1, \dots, a_{i-1}, a_i+a_i', a_{i+1}, \dots, a_n\right) \\ =\; &\Delta\left(a_1, \dots, a_{i-1}, a_i, a_{i+1}, \dots, a_n\right) \\ +\; &\Delta\left(a_1, \dots, a_{i-1}, a_i', a_{i+1}, \dots, a_n\right) \end{align}

• Second:

\begin{align} &\Delta\left(a_1, \dots, a_{i-1}, \lambda a_i, a_{i+1}, \dots, a_n\right) \\ =\; \lambda&\Delta\left(a_1, \dots, a_{i-1}, a_i, a_{i+1}, \dots, a_n\right) \end{align}

I know that the multilinear forms form a vector space over $K$. Let $W$ be that vector space.

Now I want to figure out what $\dim_K W$ is but I don't know where to start. Can you help me?

Answer 1

Hint: Define maps $\varphi_{i_1,\ldots,i_n}(v_1,\ldots,v_n) = (e_{i_1}^*v_1)\cdot (e_{i_2}^*v_2)\cdots (e_{i_n}^*v_n)$ where $\{e_i\}$ is base for $V$ and $\{e_i^*\}$ its dual base. Show that these are multilinear, linearly independent and generate your space.

Answer 2

If you have a bilinear form $\varphi\colon U\times V\to K$, you can define a map $\varphi_v\colon U\to V^*$, for each $u\in U$, by $$ \varphi_u(v)=\varphi(u,v) $$ and this is a linear map.

Conversely, if you have a linear map $f\colon U\to V^*$, you can define a bilinear map $\hat{f}\colon U\times V\to K$ by $$ \varphi(u,v)=f(u)(v) $$

This implies that the space $\operatorname{Bilin}(U\times V,K)$ of bilinear maps $U\times V\to K$ has dimension $(\dim U)(\dim V^*)=(\dim U)(\dim V)$, since it is isomorphic to $\operatorname{Lin}(U,V^*)$.

More generally, $\operatorname{Bilin}(U\times V,W)$ is isomorphic to $$ \operatorname{Lin}(U,\operatorname{Lin}(V,W)) $$ where $W$ is any vector space. You can prove this in the same fashion.

We can generalize this to get an isomorphism $$\DeclareMathOperator{\ML}{Multilin} \ML(V_1\times V_2\times\dots\times V_n,W)\to \ML(V_1\times V_2\times\dots\times V_{n-1}, \operatorname{Lin}(V_n,W))\tag{*} $$ For each $x\in V_n$ and each $\varphi\in\ML(V_1\times V_2\times\dots\times V_n,W)$, define $$ \varphi_x\colon V_1\times V_2\times V_{n-1}\to \operatorname{Lin}(V_n,W)),\qquad \varphi_x(v_1,\dots,v_{n-1})=\varphi(v_1,\dots,v_{n-1},x) $$

For $n=1$, the dimension of $\ML(V_1,W)=\operatorname{Lin}(V_1,W))= (\dim V_1)(\dim W)$, so we can conjecture that $$ \dim\ML(V_1\times V_2\times\dots\times V_n,W)= (\dim V_1)(\dim V_2)\dotsm(\dim V_n)(\dim W) $$ and prove it by induction using the isomorphism in (*).

Answer 3

Let's denote the vector space $W$ by $L^{n}(V_{m}, \mathbb{K})$, where $m = \dim V$. We claim that $\dim L^{n}(V_{m}, \mathbb{K}) = m^{n}$. Let $\{a_{i}, \dots, a_{m}\}$ a basis of $V$. If $x_{j} \in V$, let's write $$ x_{j} = x_{1j}a_{1} + x_{2j}a_{2} + \dots + x_{nj} a_{m} = \sum_{k_{j} = 1}^{m} x_{k_{j} j}a_{k_{j}}. $$ Let $\Delta \in L^{n}(V_{m}, \mathbb{K})$. Appliying linearity we have $$ \Delta(x_{1}, \dots, x_{n}) = \sum_{k_{1}, \dots, k_{n} = 1}^{m} x_{k_{1}1}x_{k_{2}2} \dots x_{k_{r}r}\Delta(a_{k_{1}}, \dots, a_{k_{n}}). $$

Let $J_{n}$ be the set $\{1, \dots, n\}$, $n \in \mathbb{N}$, and let $t: J_{n} \to J_{m}$ be the function $t(j) = k_{j}$ for all $j = 1, \dots, n.$ Let's write $$ x_{t} := x_{t(1)1} x_{t(2)2} \dots x_{t(n)n} = x_{k_{1}1}x_{k_{2}2} \dots x_{k_{r}r}, \quad a_{t} := (a_{t(1)}, \dots, a_{t(n)}) = (a_{k_{1}}, \dots, a_{k_{n}}). $$ Let $\mathcal{F}(J_{n}, J_{m})$ be the set of all functions from $J_{n}$ to $J_{m}$ (there are $m^{n}$ elements in this set). Now we can write $$ \Delta(x_{1}, \dots, x_{n}) = \sum_{t \in \mathcal{F}(J_{n}, J_{m})} x_{t} \Delta(a_{t}) $$ Observe that the functions $\mu_{t}(x_{1}, \dots, x_{r}) = x_{t}$ are multilinear forms. We shall proof the the set $\{\mu_{t}: t \in \mathcal{F}(J_{n}, J_{m})\}$ is linearly independent. In fact, if $$ \sum_{t \in \mathcal{F}(J_{n}, J_{m})} \lambda_{t} \mu_{t} = 0 $$ then, evaluating at $a_{s}$ for some $s \in \mathcal{F}(J_{n}, J_{m})$ $$ \sum_{t \in \mathcal{F}(J_{n}, J_{m})} \lambda_{t} \mu_{t} (a_{s})= \lambda_{s} = 0 \quad (\text{note that } \mu_{t}(a_{s}) = \delta_{st}). $$ We conclude that if $\Delta \in L^{n}(V_{m}, \mathbb{K})$ then $$ \Delta = \sum_{t \in \mathcal{F}(J_{n}, J_{m})} \Delta(a_{t}) \mu_{t} $$ hence $\{\mu_{t}: t \in \mathcal{F}(J_{n}, J_{m})\}$ is a basis with $m^{n}$ elements for $L^{n}(V_{m}, \mathbb{K})$.