If $v_i$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 $v_1, \ldots, v_k \in V$ are linearly dependent, proof $\det[\phi_i(v_j)] = 0.$ 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.

This question has a cousin here: If $\phi_i$s are linearly dependent, $\det [\phi_i(v_j)] = 0$ - is the proof legit?. Though my original attempt is wrong, I think the answers are indeed great.

Answer 1

I'm not sure what the $\lambda$s are above.

If the $v_k$ are linearly dependent, then for some $\alpha \neq 0$, then $\sum_j \alpha_i v_j = 0$. Without loss of generality, we may assume $\alpha_1 \neq 0$ and write $v_1 = - \sum_{j \neq 1} \frac{\alpha_j}{\alpha_1} v_j$.

Let $M$ be the matrix with entries $[M]_{ij} = \phi_i(v_j)$.

Then we have $\phi_i(v_1) = - \sum_{j \neq 1} \frac{\alpha_j}{\alpha_1} \phi_i(v_j)$, so we see that the first column of $M$ is a linear combination of the other columns, hence $\det M = 0$.

Explicitly, using the above, we can write $M = \begin{bmatrix} 0 & -\frac{\alpha_2}{\alpha_1} & -\frac{\alpha_3}{\alpha_1} & \cdots & -\frac{\alpha_n}{\alpha_1} \\ 0 & 1& 0 & \cdots & 0 \\ 0 & 0& 1 & \cdots & 0 \\ \vdots & & & \ddots& \vdots\\ 0 & 0 & \cdots & 0 & 1 \end{bmatrix} M$, from which we see that $\det M = 0$.