Rank of a matrix constructed by a dual space.

Question

Let $V$ be a vector space of a finite dimension.

Let $ B = \left \{v_1,....v_n \right \} \subseteq V$, such that $\text{dim}(\text{Span}(B)) = k$

Let $C = (\lambda_1,....\lambda_n ) \subseteq V^{*}$ be a sequence of arbitrary linear functionals.

We'll define the following $n \times n$ matrix by: $$ A = \begin{pmatrix} \lambda_1(v_1) & \cdots & \lambda_1(v_n)\\ \vdots & & \vdots\\ \lambda_n(v_1)& \cdots & \lambda_n(v_n) \end{pmatrix}$$

Prove that $rank(A) \leq k $

I can understand why $rank(A) \leq k $, but I really trouble myself finding the correct way to put it down. i'd love to get some help in this manner!

Answer 1

We can solve it via a well-known conclusion $$ \text{rank} (AB) \le \min \{\text{rank} (A), \text{rank} (B)\}$$

In order to reveal this, we just need to transform the original problem as follows

Considering that $\dim \text{Span} (B) =k$, we write $$ \tilde{v}_i = (v_{i1},v_{i2},\cdots,v_{ik}) \quad\forall \, 1\le i \le n $$ Then the elements in the dual space can be regarded as dual vectors $$ \tilde{\lambda}_i = (\lambda_{i1},\lambda_{i2},\cdots,\lambda_{ik}) \quad\forall \, 1\le i \le n $$ and we have $$\lambda_i(v_j) = \tilde{\lambda}_i \cdot \tilde{v}_j$$

Thus $$ A= \begin{pmatrix} \tilde{\lambda}_1 \\ \vdots \\ \tilde{\lambda}_n \end{pmatrix} \left( \tilde{v}_1^T,\cdots, \tilde{v}_n^T \right) $$ Then the conclusion mentioned above yields that $\text{rank}(A) \le \text{rank} \left( \tilde{v}_1^T,\cdots, \tilde{v}_n^T \right)= k$

Answer 2

If $\text{rank}\ A>k$ then for some $l>k,$ an $l\times l$ cofactor of $A$ has non zero determinant. Without loss of generality, assume that this cofactor, call it $A_1$, is the upper left-hand corner of $A.$ Then $A_1$ corresponds to an invertible linear transformation $T:\mathbb R^l\to \mathbb R^l$.

If $\{v_1,\cdots, v_l\}$ were not linearly independent, then there is a $\vec c=(c_1,\cdots, c_l);\ c_j\neq 0$ for some $1\le j\le l$ such that $\sum^l_{i=1}c_i v_i=0.$ But then, from the linearity of the $(\lambda_i)^k_{i=1},$ it follows that $A_1\vec c=0$, which contradicts the invertibility of $T$. And since $l>k,$ this implies that $\text{dim span}\ B>k$, and we have a contradiction.