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, 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.

  • $\begingroup$ Why is $\phi_i(v_j) = \delta_{ij}$? Did you omit some assumption from the problem? $\endgroup$ – Magdiragdag Jun 25 '13 at 21:26
  • $\begingroup$ In fact I meant some isomorphism with $\delta_{ij}$. $\endgroup$ – WishingFish Jun 25 '13 at 21:27
  • $\begingroup$ I don't think I left some assumption from the problem - just my proof is not well-rounded.. $\endgroup$ – WishingFish Jun 25 '13 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. $\endgroup$ – Magdiragdag Jun 25 '13 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 Jun 25 '13 at 21:37

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


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...)

  • $\begingroup$ ........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 $\endgroup$ – WishingFish Jun 25 '13 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 Jun 26 '13 at 5:41
  • $\begingroup$ -_- Thanks... Now I am more or less figured out I was completely wrong... Thanks.. $\endgroup$ – WishingFish Jun 26 '13 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 $\phi^i(E_j) = \delta^i_j$." $\endgroup$ – WishingFish Jun 26 '13 at 17:29

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.