# ${\Lambda}_k(V)$ is spanned by a basis $\mathcal{B}$

Let $V$ be a vector space over $\mathbb{R}$ and let ${\Lambda}_k(V)$ be the set of all the multilinear-alternating functions from $\prod_{i = 1}^k V$ to $\mathbb{R}$. We are going to suppose that $k \leq n = \dim V$ and $\{v_1 , \ldots , v_n\}$ and $\{v_1^* , \ldots , v_n^*\}$ are basis for $V$ and $V^*$ respectively. Fix $f \in {\Lambda}_k(V)$ and I want to show that $$f = \sum_{1\leq i_1 < \ldots < i_k\leq n} {\lambda}_{i_1 , \ldots , i_k} \left(\bigwedge_{j = 1}^k v_{i_j}^*\right)\mbox{,}$$ being ${\lambda}_{i_1 , \ldots , i_k} = f(v_{i_1} , \ldots , v_{i_k})$, which implies that ${\Lambda}_k(V)$ is spanned by $$\mathcal{B} = {\left\{\bigwedge_{j = 1}^k v_{i_j}\right\}}_{1\leq i_1 < \ldots < i_k\leq n}\mbox{.}$$

In http://www.maths.adelaide.edu.au/michael.murray/dg_hons/node25.html can be found that it is suggested to develop the right member of last equation and it's my attemp about. Fixed $u_j \in V$, $k = 1 , \ldots , k$, I am trying to prove that $$\sum_{1\leq i_1 < \ldots < i_k\leq n} {\lambda}_{i_1 , \ldots , i_k} \left(\bigwedge_{j = 1}^k v_{i_j}^*\right)(u_1 , \ldots , u_k) = f(u_1 , \ldots , u_k)\mbox{.}$$ Well, at first we need to express $u_k$ through the vectors $v_i$, $i = 1 , \ldots , n$. We are going to suppose that $$u_j = \sum_{i = 1}^n {\lambda}_i^j v_i \quad \mbox{ for all } \quad j = 1 , \ldots , k\mbox{.}$$ Therefore, $$\sum_{{i_1 , \ldots , i_k \in \{1 , \ldots , n\}} \atop {i_1 < \ldots < i_k}} f(v_{i_1} , \ldots , v_{i_k}) \left(\bigwedge_{l = 1}^k v_{i_l}^*\right)(u_1 , \ldots , u_k) =$$ $$= \sum_{{i_1 , \ldots , i_k \in \{1 , \ldots , n\}} \atop {i_1 < \ldots < i_k}} f(v_{i_1} , \ldots , v_{i_k}) \left(\sum_{j_1 , \ldots , j_k \in \{1 , \ldots , n\}} \left(\prod_{l = 1}^k {\lambda}_{j_l}^l\right) \left(\bigwedge_{l = 1}^k v_{i_l}^*\right)(u_1 , \ldots , u_k)\right) =$$ $$= \frac{1}{k !} \sum_{{i_1 , \ldots , i_k \in \{1 , \ldots , n\}} \atop {i_1 < \ldots < i_k}} f(v_{i_1} , \ldots , v_{i_k}) \left(\sum_{j_1 , \ldots , j_k \in \{1 , \ldots , n\}} \left(\prod_{l = 1}^k {\lambda}_{j_l}^l\right) \left(\sum_{\sigma \in G_{j_1 , \ldots , j_k}} sgn(\sigma) \left(\prod_{l = 1}^k {\delta}_{i_l , \sigma(j_l)}\right)\right)\right) =$$ $$= \frac{1}{k !} \sum_{{i_1 , \ldots , i_k \in \{1 , \ldots , n\}} \atop {i_1 < \ldots < i_k}} f(v_{i_1} , \ldots , v_{i_k}) \left(\sum_{j_1 , \ldots , j_k \in \{1 , \ldots , n\} \atop \sigma \in S_k} sgn(\sigma)\left(\prod_{l = 1}^k {\lambda}_{j_l}^l {\delta}_{i_l , j_{\sigma(l)}}\right)\right)\mbox{,}$$ where $G_{j_1 , \ldots , j_k} \cong S_k$, being $\psi : S_k \to G_{j_1 , \ldots , j_k}$, given by $\psi : l \mapsto j_l$ an isomorphism ($G_{j_1 , \ldots , j_k}$ is essentially $S_k$). I do not know how I can continue since here. Thank you very much in advance.

• You can use 1\leq i_1 < \ldots < i_k\leq n for the result $1\leq i_1 < \ldots < i_k\leq n$, which makes just one line and a less tiny output. – Arnaud Mortier May 24 '18 at 0:19
• Thank you by your suggestion. I have just modified the text. – joseabp91 May 24 '18 at 0:23

I believe that the step $$\sum_{1\leq i_1 < \ldots < i_k\leq n} {\lambda}_{i_1 , \ldots , i_k} \left(\bigwedge_{j = 1}^k v_{i_j}^*\right)(u_1 , \ldots , u_k) = \sum_{1\leq i_1 < \ldots < i_k\leq n} {\lambda}_{i_1 , \ldots , i_k} \left(\prod_{j = 1}^k v_{i_j}^*(u_j)\right)$$ is wrong.

If this were true, for instance you would have $$e_1^*\wedge e_2^*(e_2,e_1)=e_1^*(e_2)e_2^*(e_1)=0$$ instead of the expected $$e_1^*\wedge e_2^*(e_2,e_1)=-e_1^*\wedge e_2^*(e_1,e_2)=-1.$$

So what you need to do is split the $u_i$ before feeding them to the $v_{i_j}^*$.

The correct continuation from there is $$\sum {\lambda}_{i_1 , \ldots , i_k} \left(\bigwedge_{j = 1}^k v_{i_j}^*\right)(u_1 , \ldots , u_k) =\sum {\lambda}_{i_1 , \ldots , i_k} \left(\bigwedge_{j = 1}^k v_{i_j}^*\right)\left(\sum_{i = 1}^n {\lambda}_i^1 v_i , \ldots , \sum_{i = 1}^n {\lambda}_i^n v_i\right)$$ Then use multilinearity and consider all ways to select one $v_i$ at a time so as to have exactly the list $\{v_{i_1},\ldots, v_{i_n}\}$ in the end (with a possible minus sign depending on the parity of the permutation).

Note, however, that you don't have to actually do this. Only observe that it can be done, and that it can be done in the exact same manner for $f$ since it is multilinear and alternate as well, giving the same coefficients that depend on the matrix $[\lambda_i^j]$.

• How can I split the $u_i$? – joseabp91 May 24 '18 at 0:47
• @joseabp91 You did it already, you defined the coordinates $\lambda_i^j$. – Arnaud Mortier May 24 '18 at 0:49
• So, how can I prove the equality? I do not understand. – joseabp91 May 24 '18 at 0:58
• @joseabp91 I edited. – Arnaud Mortier May 24 '18 at 1:18
• I have just developed the right member of your equation using multinilearity but I have obtained the same than before. – joseabp91 May 24 '18 at 9:47