1
$\begingroup$

From (Italian, very nice book):"Lezioni di Geometria Analitica e Proiettiva" by Beltrametti, CArletti, Gallarati, Bragadin (pag. 21):

Let $K$ a field, $V:= K^{n+1}$ and let $e_1,\ldots, e_{n+1}$ a base (canonical or not) of $V$. Let $W\subset K^{n+1}$ a $K$-vectorial subspace with dimension $r+1$, and let $v_1,\ldots, v_{r+1}$ a base of $W$, with

$v_m= a^1_m\cdot e_1 + \ldots a^{n+1}_m\cdot e_{n+1}$ for $1\leq m\leq r+1$

Let $M$ the matrix with ($n+1$) row's:

$x_1, a^1_1\ldots, a^1_{r+1} $

$x_2, a^2_1\ldots, a^2_{r+1} $

$\ldots, \ldots, \ldots$

$\ldots, \ldots, \ldots$

$x_{n+1}, a^{n+1}_1, \ldots a^{n+1}$

(the last element is $a^{n+1}_{r+1}$)

The book assert (mentioning Kronecker theorem) that

the $r+2$-minor's of $M$ (these are $\binom{n+1}{r+2}$)

considered as linear forms (grade 1 homogeneous polynomial) on variables $x_1,\ldots, x_{n+1}$

are linearly dependent, and there are $n-r$ (and no more) linearly independent $r+2$-minors.

Is this true?

How to prove this?

$\endgroup$
5
$\begingroup$

Consider the map $\phi: V\to K^{M}$ where $M=\binom{n+1}{r+1}$, which sends a vector $y\in V$ to the $M-$tuple of minors (ordered as you wish) of the matrix $$A_y=(y\ \vert\ v_1\ \vert\;\cdots\;\vert\ v_{r+1})$$ then $y\in W$ if and only if $\phi(y)=0$, because $y\in W$ iff it is linearly dependent on $\{v_1,\ldots, v_{r+1}\}$, fact that happens iff $\mathrm{rk}A_y=r+1$ iff all the $(r+2)$-minors of $A_y$ vanish. Therefore, $W=\ker \phi$. Now, write $\phi=(\phi_1,\ldots, \phi_M)$, with $\phi_j\in V^*$. We have that $W=\{\phi_1=\ldots=\phi_M=0\}=\left(\mathrm{Span}\{\phi_1,\ldots,\phi_M\}\right)^0$, but $\dim W=r+1$, so $\dim\mathrm{Span}\{\phi_1,\ldots,\phi_M\}=(n+1)-(r+1)=n-r$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, only I think all but this (simply) way . Grazie, ho pensato di tutto, tranne che questa via. $\endgroup$ – Buschi Sergio Dec 17 '12 at 8:38

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.