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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.

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  • $\begingroup$ Why is $\phi_i(v_j) = \delta_{ij}$? Did you omit some assumption from the problem? $\endgroup$ Commented Jun 25, 2013 at 21:26
  • $\begingroup$ In fact I meant some isomorphism with $\delta_{ij}$. $\endgroup$ Commented Jun 25, 2013 at 21:27
  • $\begingroup$ I don't think I left some assumption from the problem - just my proof is not well-rounded.. $\endgroup$ Commented Jun 25, 2013 at 21:28
  • $\begingroup$ I'm sorry, you've lost me there. $\phi_i(v_j)$ is some element of the underlying field. Because of the equality between the matrices that you write, I assumed that you $\phi_i(v_j) = 1$ if $i = j$ and $0$ if $i \neq j$. Or do you just denote $\phi_i(v_j)$ by $\delta_{ij}$? Fair enough, but then you need some argument for the equality between the matrices. $\endgroup$ Commented Jun 25, 2013 at 21:35
  • $\begingroup$ Either I'm misunderstanding something big time or something here makes no sense: The $\;\phi_i\;$ are linear functionals on $\,\;V\;$ since we're given that $\,\phi_i\in V^*\,$, and thus for any $\,u\in V\;,\;\;\phi(u)\in\Bbb F\;,\;\;\Bbb F=$ the definition field. How is it possible then that $\,\phi(v_j)\,$ is... a matrix ?? And if we're supposed to take the matrix $\,A:=\left(\phi_i(v_j)\right)\;$ and then its determinant then it must be $\,p=k\,$ ... $\endgroup$
    – DonAntonio
    Commented Jun 25, 2013 at 21:37

2 Answers 2

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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...)

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  • $\begingroup$ ........I think I did the problem as if $v_i$s are linearly independent................!!!!! In this case, is my original proof correct? Thank you.l $\endgroup$ Commented Jun 25, 2013 at 22:59
  • $\begingroup$ I can't see where you assume the $\,v_i'$s are lin. ind. Anyway, assuming the more or less obvious things, it looks like somehow you're assuming $\,\phi_i(v_j)=\delta_{ij}=$ Kronecker's delta, god knows why or how. I think your explanation lacks lots of details, the above being one of them. $\endgroup$
    – DonAntonio
    Commented Jun 26, 2013 at 5:41
  • $\begingroup$ -_- Thanks... Now I am more or less figured out I was completely wrong... Thanks.. $\endgroup$ Commented Jun 26, 2013 at 16:48
  • $\begingroup$ The wrong idea comes from this: "If $(E_1, \dots, E_n)$ is a basis for $V$, we let $(\phi^1, \ldots, \phi^n)$ denote the corresponding dual basis for $V^*$, defiend by $\phi^i(E_j) = \delta^i_j$." $\endgroup$ Commented Jun 26, 2013 at 17:29
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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.

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