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Jan 17, 2012 at 0:34 comment added darij grinberg ... a strictly decreasing chain of subspaces of a finite-dimensional space, dimensions must go down by at least $1$ in each step. This is impossible for this chain, since $\dim W_{d+1} = d-1$ and $\dim\left(W_{d+1}\cap W_d\cap ...\cap W_1\right)\geq 0$.
Jan 17, 2012 at 0:33 comment added darij grinberg Something seems wrong with the question: (1) alone is impossible to fulfill! To see this, notice that, for every $i\in\left\lbrace 1,2,...d\right\rbrace$, the space $W_{d+1}\cap W_d\cap ...\cap W_i$ is a proper subspace of $W_{d+1}\cap W_d\cap ...\cap W_{i+1}$ (because otherwise, $W_i + W_{i+1}\cap ...\cap W_{d+1}$ would be $\subseteq W_i$ and thus couldn't be the whole $V$, contradicting (1)). This yields that $W_{d+1} \supset W_{d+1}\cap W_d \supset W_{d+1}\cap W_d\cap W_{d-1} \supset ... \supset W_{d+1}\cap W_d \cap ...\cap W_1$, where all inclusions are proper. But in ...
Jan 16, 2012 at 21:47 history edited Alireza Abdollahi CC BY-SA 3.0
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Jan 16, 2012 at 21:41 history edited Alireza Abdollahi CC BY-SA 3.0
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Jan 16, 2012 at 21:31 history asked Alireza Abdollahi CC BY-SA 3.0