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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 dimention $r+1$, and let $v_1,\ldots, v_{r+1}$ a base of $W$, with

$v_m= a^1_m\cdot e_1 + \ldots a^{n+1}_m\cdot e_n$ 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 $x_1,\ldots, x_n$

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

Is this true?

How to prove this?

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1 Answer 1

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 $\dim\mathrm{Span}\{\phi_1,\ldots,\phi_M\}=(n+1)-(r+1)=n-r$.

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Thank you, only I think all but this (simply) way . Grazie, ho pensato di tutto, tranne che questa via. –  Buschi Sergio Dec 17 '12 at 8:38
    
You are welcome! Prego. –  Samuele Dec 17 '12 at 11:44
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