In linear algebra text(Hoffman), it says "In a vector space of dimension $n$, a subspace of dimension $n-1$ is called hyperspace".
The problem is the following theorem.
If $W$ is a $k$-dimensional subspace of an $n$-dimensional vector space V, then W is the intersection of $(n-k)$ hyperspaces in V.
I don't understand what $(n-k)$ hyperspaces is. The definition of hyperspace is $n-1$ subspace. Does it mean $(n-k-1)$ subspace?
I truly know the fact that where $W^0$ is the annihilator of $W$, " dim $W^0$ + dim$W$ = dim $V$ " holds. And by this formula, W is exactly the set of vectors $\alpha$ such that $f_i(\alpha) = 0$, $i = k+1,...,n $. if $\alpha_j (j=1,...n)$ is basis of V and $f_i(\alpha_j)=\delta_{ij}$.
The text says "In case $k=n-1$, $W$ is the null space of $f_n$, so it is proved.", but i can't understand why $k=n-1$ holds and why it is proved(actually i don't understand the proof).
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$\begingroup$It means that there are hyperspaces $H_1, H_2, \dots, H_{n-k}$ such that $H_1\cap H_2\cap \dots\cap H_{n-k} = W$. That is, the number of hyperspaces needed is $n - k$.
In the case of $k = n - 1$, $W$ is itself a hyperspace so you only need $1 = n - (n - 1)$ hyperspaces.
$\endgroup$ $\begingroup$The definition of hyperspace is an $(n-1)$-dimensional subspace.
For example, if $V$ is $(n=\ )\ 2$-dimensional, then hyperspaces are $(n-1=\ )\ 1$-dimensional subspaces (lines), and he claims that the $(k=\ )\ 0$-dimensional subspaces (points) are the intersection of $n-k = 2-0 = 2$ lines which should seem familiar.
If you want it to be $(n-k-1)$-dimensional subspaces, then here you would need every point to be a line!
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