# If $v_i$s are linearly dependent, $\det [\phi_i{v_j}] = 0$ - is the proof legit?

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.

• I presume that the question should read $\phi_k \in V^*$? – copper.hat Jun 26 '13 at 2:09
• Yeah, thanks so much for pointing it out... – WishingFish Jun 26 '13 at 2:14
• This question shows most of the other question's features: why do you keep on assuming, apparently, that $\,\phi_i(v_j)=\delta_{ij}=$Kronecker's delta? This does not have to be true in the general case. – DonAntonio Jun 26 '13 at 6:03

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 $\det M = 0$.

• $\lambda$s are the coefficient of linear combination. I guess is your $\alpha$s.. -) – WishingFish Jun 26 '13 at 2:25
• Ah, basically we have the same thing except transposed. You need to multiply the bottom matrix in your question by the top matrix for the equality to be correct (as in mine above, where $M$ appears on both the right and left). – copper.hat Jun 26 '13 at 2:31
• Actually I think my solution is wronger than that... The bottom matrix was intended to be the first matrix after carrying out the values of $\delta$. But I don't think the employment of $\delta$ is correct... – WishingFish Jun 26 '13 at 2:33
• Well, you have the right idea. The rest are details. – copper.hat Jun 26 '13 at 2:38
• =) Thank you very much for your encouragement! – WishingFish Jun 26 '13 at 2:38