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What I don't understand here is why is $h(\alpha)=0$ for all $\alpha$ in $N_{k}$. Is there a typo?

In case there is not, could someone please detail that last step please?

shooting-squirrel
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    By construction, the restriction of $h$ to $N_k$ is $g' - \sum\limits_{i=1}^{k-1}f_i'$ - which is $0$, by construction. – Daniel Fischer Sep 07 '13 at 21:46
  • Is it that simple? huh?(scratching my head...) – shooting-squirrel Sep 07 '13 at 21:55
  • Does the author use the induction ipothesis: $g=\sum^{k-1}{i=1}c{i}f_{i}$ ? If not what is the induction ipothesis? – shooting-squirrel Sep 07 '13 at 22:12
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    The induction hypothesis is that for all vector spaces $V$, and all systems $(\mu,{\lambda_i : 1 \leqslant i \leqslant r})$ of linear functionals on $V$, with $r < k$, and $\mathcal{N}(\mu) \supset \bigcap\limits_{i=1}^r \mathcal{N}(\lambda_i)$, you can write $\mu$ as a linear combination of the $\lambda_i$. The induction hypothesis is applied to the space $V = N_k$, and the system $(g', { f_i' : 1 \leqslant i < k})$. – Daniel Fischer Sep 07 '13 at 22:17
  • I don't understand the notation you use, could you please use a simpler notation? Or explain for example, what do you mean by systems of linear functionals, or what do you mean by the big fancy character that looks like N? – shooting-squirrel Sep 07 '13 at 22:21
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    The $\mathcal{N}(f)$ denotes the null space, or kernel, of $f$. By system, I mean the kind of data the theorem is concerned with. You have a finite set of linear functionals (the $f_i$ resp $\lambda_i$), and another linear functional ($g$ or $\mu$) that vanishes on the intersection of the null spaces of the $f_i$ ($\lambda_i$). – Daniel Fischer Sep 07 '13 at 22:26
  • Do you mean "vanishes off" or "vanishes of"? – shooting-squirrel Sep 07 '13 at 22:30
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    I meant "vanishes on". Fixed now. – Daniel Fischer Sep 07 '13 at 22:31
  • I think I understand. In the case it's not too much to ask from you, what's the point of the sentence : "Furthermore, if $\alpha$ is a vector in $N_{k}$ and $f'{i}=0$, $i=1, ..., k-1$, then $\alpha$ is in the intersection of the $N{i}$'s (i from 1 to k-1) and so $g'(\alpha)=0$."? – shooting-squirrel Sep 07 '13 at 22:39
  • It says that the system of the restrictions of $g$ and the $f_i$ to $N_k$ satisfies the conditions of the theorem. Without restricting to $N_k$, we don't know that $g$ vanishes on $\bigcap\limits_{i=1}^{k-1} N_i$ - in general, it doesn't. But $\bigcap\limits_{i=1}^{k-1} N_i' = \bigcap\limits_{i=1}^k N_i$, so the restriction of $g$ to $N_k$ vanishes on that. – Daniel Fischer Sep 07 '13 at 22:45
  • Got it, thanks a lot! – shooting-squirrel Sep 07 '13 at 23:06

1 Answers1

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Answered by Daniel Fischer in a comment

By construction, the restriction of $h$ to $N_k$ is $g' - \sum\limits_{i=1}^{k-1}f_i'$ - which is $0$, by construction.

Additional details

The induction hypothesis is that for all vector spaces $V$, and all systems $(\mu,\{\lambda_i : 1 \leqslant i \leqslant r\})$ of linear functionals on $V$, with $r < k$, and $\mathcal{N}(\mu) \supset \bigcap\limits_{i=1}^r \mathcal{N}(\lambda_i)$, you can write $\mu$ as a linear combination of the $\lambda_i$. The induction hypothesis is applied to the space $V = N_k$, and the system $(g', \{ f_i' : 1 \leqslant i < k\})$.

Here $\mathcal{N}(f)$ denotes the null space, or kernel, of $f$. By system, I mean the kind of data the theorem is concerned with. You have a finite set of linear functionals (the $f_i$ resp $\lambda_i$), and another linear functional ($g$ or $\mu$) that vanishes on the intersection of the null spaces of the $f_i$ ($\lambda_i$).

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