Multilinear Map of a Determinant: Basis Extension from Subspace

Question

Let $V$ be an $n$-dimensional $K$-vector space, $U ⊆ V$ an $r$-dimensional linear subspace and $x_{r + 1},\dots , x_{n}$ a fixed System of vectors in $V$. For a determinant function $\Delta: V^n \rightarrow K$ (i.e. an alternating $n$-multilinear form) show that through $$ \Delta_U (a_1,\dots, a_r):= \Delta (a_1,\dots, a_r, x_{r + 1},\dots, x_{n})$$ a determinant function $\Delta_U:U^r \rightarrow K$ can be defined. When is $\Delta_U$ non-trivial?

Answer 1

We want to show that $\Delta_U: U \rightarrow K, \Delta_U(a_1,\dots, a_r) := \Delta(a1,...,a_r,x_{r+1},...,x_n)$ is a determinant function.

Proof: The properties for the determinant function are inherited from $\Delta$:

$$\Delta_U(a_1,\dots, a_{i1}+a_{i2},...,a_r)= \Delta(a_1,\dots, a_{i1}+a_{i2},...,a_r,x_{r+1},...,x_n)=\Delta(a_1,\dots, a_{i1},...,a_r,x_{r+1},...,x_n)+\Delta(a_1,\dots,a_{i2},...,a_r,x_{r+1},...,x_n)=\Delta_U(a_1,\dots, a_{i1},...,a_r)+\Delta_U(a_1,\dots,a_{i2},...,a_r)$$

$$\Delta_U(a_1,\dots, \lambda a_i,...,a_r)=\Delta(a_1,\dots, \lambda a_i,...,a_r,x_{r+1},...,x_n)=\lambda \Delta(a_1,\dots, a_i,...,a_r,x_{r+1},...,x_n)=\lambda \Delta_U(a_1,\dots,a_i,...,a_r)$$

$$\exists i \ne j: a_i = a_j \Rightarrow \Delta_U(a_1,\dots,...,a_r)= \Delta(a_1,\dots,a_r,,x_{r+1},...,x_n)=0$$

When is $\Delta_U$ non-trivial? First consider when $\Delta_U$ is trivial. We get three cases:

• $\Delta$ is trivial

• $\exists \lambda_{r+1},...,\lambda_n \in K: \sum^n_{i=r+1} \lambda_i x_i \in U \ \{0\}$ (because then $(a_1,..., a_r, x_{r + 1},..., x_n)$ is a linearly dependent system for all $a_1,. . . , a_r \in U$) That is: $\exists u \in U \ \{0\}: u \in L(x_{r + 1},..., x_n)$

• $(x_{r + 1},..., x_n)$ forms a linearly dependent system. So $\Delta_U$ is non-trivial if $$\Delta\operatorname{not trivial} \land \space \forall \space u \in U \ \{0\}: u \notin L(x_{r + 1},..., x_n) \land (x_{r + 1},..., x_n)$$

is linearly independent.

(You could also summarize the last two statements as follows: $(x_{r + 1},..., x_n)$ forms a Basis of a complementary space from $U$ in $V$).