If $\phi_i$s are linearly dependent, $\det [\phi_i(v_j)] = 0$ - is the proof legit?

Question

Given $v_1, \ldots, v_k \in V$ and $\phi_1, \ldots, \phi_k \in V^*$. If $\phi_1, \ldots, \phi_k \in V^*$ are linearly dependent, prove that $\det[\phi_i(v_j)] = 0.$ Here $k$ is the dimension of $V$, but I need to show this also works for a subspace with dimension $p$.

Assume $v_r$ is a linear combination of the others, as without loss of generality we consider in terms of basis. Denoting the matrix whose $i$th row and $j$th column is $\phi_i(v_j)$ to be $[\phi_i(v_j)]$. Then

$$[\phi_i(v_j)]= \pmatrix{ \begin{array}{cccccc} \delta_{11}&\delta_{12} & \ldots & \lambda_1 \delta_{11} + \cdots + \hat \lambda_r \delta_{1r} + \cdots + \lambda_p \delta_{1p} & \cdots & \delta_{1p}\\ \delta_{21}&\delta_{22} & \ldots & \lambda_1 \delta_{21} + \cdots + \hat \lambda_r \delta_{2r} + \cdots + \lambda_p \delta_{2p} &\cdots & \delta_{2p}\\ \vdots &\vdots & \ddots &\vdots & \ddots & \vdots\\ \delta_{r1}&\delta_{r2} & \ldots & \lambda_1 \delta_{r1} + \cdots + \hat \lambda_r \delta_{rr} + \cdots + \lambda_p \delta_{rp} &\cdots & \delta_{rp}\\ \vdots &\vdots & \ddots &\vdots & \ddots & \vdots\\ \delta_{p1}&\delta_{p2} & \ldots & \lambda_1 \delta_{p1} + \cdots + \hat \lambda_r \delta_{pr} + \cdots + \lambda_p \delta_{pp}&\cdots & \delta_{pp}\\ \end{array} } =\left( \begin{array}{cccccc} 1&0 & \ldots & \lambda_1 & \cdots & 0\\ 0&1 & \ldots & \lambda_2 & \cdots & 0\\ \vdots &\vdots & \ddots & \vdots &\ddots & \vdots\\ 0&0 & \ldots & 0 & \cdots & 0\\ \vdots &\vdots & \ddots & \vdots &\ddots & \vdots\\ 0&0 & \ldots & \lambda_p & \cdots & 1\\ \end{array} \right). $$ Hence, the determinant of $\phi_i(v_j)$ is zero.

Answer 1

After the comments clearing up stuff and the editing of the question we can now try to answer: since $\,\{\phi_1,\ldots,\phi_k\}\,$ is linearly dependent there exists a first element which is a linear combination of the preceeding ones, say

$$\phi_r=\sum_{n=1}^{r-1}a_n\phi_n\;,\;\;a_n\in\Bbb F$$

From here we get that for every $\;i\;,\;\;1\le i\le k\;$ , we have

$$\phi_r(v_i)=\sum_{n=1}^{r-1}a_n\phi_n(v_i)$$

Can you see then how the $\;r-$th row is a linear combination of the first $\,r-1\,$ first rows in the matrix $\,\left(\phi_i(v_j)\right)\;$? This, of course, means the determinant is zero.

I think you had the right idea but you messed things up with all those deltas and lambdas (=my $\,a_i'$s...)

Answer 2

Please let me first try to phrase the question that I believe the OP is trying to ask; then with some care his proof will actually work.

Proposition. Let $V$ be a vectorspace over some field $K$. Take $v_1, \dots, v_p \in V$ and $\phi_1, \dots, v_p \in V^*$. Assume that $\phi_1, \dots, \phi_p$ are linearly dependent. Then $\det(\phi_i(v_j)) = 0$.

Proof. Let's say $\phi_k$ can be expressed as a linear combination of the other $\phi_i$; say $\phi_k = \lambda_1 \phi_1 + \dots \lambda_{k-1} \phi_{k-1} + \lambda_{k+1} \phi_{k+1} + \dots + \lambda_{p} \phi_{p}$ for certain $\lambda_i \in K$.

Write $\delta_{ij} = \phi_i(v_j)$. Then $$(\phi_i(v_j)) = \pmatrix{ \begin{array}{cccccc} \delta_{11}&\delta_{12} & \ldots & \lambda_1 \delta_{11} + \cdots + \hat \lambda_k \delta_{1k} + \cdots + \lambda_p \delta_{1p} & \cdots & \delta_{1p}\\ \delta_{21}&\delta_{22} & \ldots & \lambda_1 \delta_{21} + \cdots + \hat \lambda_k \delta_{2k} + \cdots + \lambda_p \delta_{2p} &\cdots & \delta_{2p}\\ \vdots &\vdots & \ddots &\vdots & \ddots & \vdots\\ \delta_{p1}&\delta_{p2} & \ldots & \lambda_1 \delta_{p1} + \cdots + \hat \lambda_k \delta_{pk} + \cdots + \lambda_p \delta_{pp}&\cdots & \delta_{pp}\\ \end{array} }$$ Subtracting, for each $i \neq k$, the $i$th column $\lambda_i$ times from the $k$th column, results in the matrix $$\pmatrix{ \begin{array}{cccccc} \delta_{11}&\delta_{12} & \ldots & 0 & \cdots & \delta_{1p}\\ \delta_{21}&\delta_{22} & \ldots & 0 &\cdots & \delta_{2p}\\ \vdots &\vdots & \ddots &\vdots & \ddots & \vdots\\ \delta_{p1}&\delta_{p2} & \ldots & 0&\cdots & \delta_{pp}\\ \end{array} }$$ This matrix has determinant $0$, and hence $\det(\phi_i(v_j)) = 0$ as well.