- #1

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Let ##L## be an ##n##-dimensional space. For each ##p##-vector ##\alpha\neq0## we let ##M_\alpha## be the subspace of ##L## consisting of all vectors ##\sigma## satisfying ##\alpha\wedge\sigma=0##. Prove that ##\dim(M_\alpha)\leq p##. Prove also that ##\dim(M_\alpha)=p## if and only if ##\alpha=\sigma^1\wedge\cdots\wedge\sigma^p## where ##\sigma^1,\ldots,\sigma^p## are vectors in ##L##.

(I wrote ##\sigma^1,\ldots## above, but Flanders wrote ##\sigma_1,\ldots## in this exercise. Weird: he uses upper indices in the theory sections.)

My question is this: Isn't it false that if ##\alpha=\sigma^1\wedge\cdots\wedge\sigma^p## where ##\sigma^1,\ldots,\sigma^p## are vectors in ##L##, then ##\dim(M_\alpha)=p##?