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I'm reading Hoffman and Kunze's Linear Algebra and on page 102 of the 2nd edition, they state and prove the following corollary:

Corollary. If $W$ is a $k$-dimensional subspace of an $n$-dimensional space $V$, then $W$ is the intersection of $(n-k)$ hyperspaces in $V$.

Proof. This is a corollary of the proof of Theorem 16 rather than its statement. In the notation of the proof, $W$ is exactly the set of vectors such that $f_i(\alpha) = 0$, $i = k+1,\dots,n$. In case $k = n-1$, $W$ is the null space of $f_n$.

It seems they want to prove $W=\bigcap_{i=k+1}^n \ker f_i$. I understood why $W\subset \ker \bigcap_{i=k+1}^n f_i$, I'm having problems with the converse.

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    $\begingroup$ You might need to add a little context to this. What are the $f_i$? And have you looked in the proof of Theorem 16, it seems that this is part of the argument used there. $\endgroup$
    – SamM
    May 12 '16 at 7:31
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Suppose $a\in \cap_{i=k+1}^n \ker f_i$. Then the $f_i$ also span $(\text{span}(W\cup \{a\}))^\circ$, giving us $\dim W^\circ= \dim (\text{span}(W\cup \{a\}))^\circ$. Hence $$\dim W= \dim \text{span}(W\cup \{a\})\implies a\in W\implies \cap_{i=k+1}^n \ker f_i\subseteq W$$

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  • $\begingroup$ I didn't understand why the $f_i$ span $(W\cup\{a\})^\circ$. Thank you! $\endgroup$
    – user42912
    May 12 '16 at 12:34
  • $\begingroup$ Well if $a\in \cap_{i=k+1}^n \ker f_i$, then any $g\in (\text{span}(W\cup \{a\}))^\circ$ can be expressed as $g=\sum_{i=k+1}^n g(a_i)f_i$. $\endgroup$ May 12 '16 at 13:06
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Suppose $\alpha\in \cap_{i=k+1}^n \ker f_i$. Using the notation of the book and by (3-14) on page 99, we know that $\alpha=\sum_{i=1}^nf_i(\alpha)\alpha_i$. Therefore $\alpha=\sum_{i=1}^kf_i(\alpha)\alpha_i$, so $\alpha\in W$.

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