If $W_1$ and $W_2$ are subspaces of a finite-dimensional vector space, then $W_1=W_2$ if and only if $W_1^0=W_2^0$.
My attempt: Suppose $W_1=W_2$. Then clearly $W_1^0=W_2^0$. Conversely, suppose $W_1^0=W_2^0$. It’s easy to check $W_1^0,W_2^0\leq V^*$. So $\mathrm{dim}(W_1^0)= \mathrm{dim}(W_2^0)$. By theorem 16 section 3.5, $\mathrm{dim}(W_1)+ \mathrm{dim}(W_1^0)$ $= \mathrm{dim}(V)$ $= \mathrm{dim}(W_2)+ \mathrm{dim}(W_2^0)$. So $\mathrm{dim}(W_1)= \mathrm{dim}(W_2)=k$. By theorem 16 corollary 1 section 3.5, $W_1$ $= \bigcap_{k\lt i\leq n}N_{f_i}$ $=W_2$. Hence $W_1=W_2$. Is my proof correct?
Hoffman’s proof: If $W_1 = W_2$,then of course $W_1^0= W_2^0$.If $W_1\neq W_2$,then one of the two subspaces contains a vector which is not in the other. Suppose there is a vector $\alpha$ which is in $W_2$ but not in $W_1$. By the previous corollaries (or the proof of Theorem 16) there is a linear functional $f$ such that $f(\beta) =0$ for all $\beta$ in $W_1$, but $f(\alpha)\neq 0$.Then $f$ is in $W_1^0$ but not in $W_2^0$ and $W_1^0\neq W_2^0$.
Question: I don’t understand following step of Hoffman’s proof, existence of linear functional $f$ such that $f(\beta)=0$, $\forall \beta \in W_1$, but $f(\alpha)\neq 0$.
Edit: My proof is wrong, reason is quite subtle.
Best Answer
Your attempted proof is at best highly incomplete. The hyperplanes $N_{f_i}$ ($k+1\le i\le n$) in Corollary 1 depend upon the subspace $W$. The Corollary does not tell you that any two $k$-dimensional subspaces $W_1$ and $W_2$ are the intersection of the same $n-k$ hyperplanes in $V$. So how do we know that $W_1=W_2$?
In Hoffman and Kunze's proof, if we write $W_1=\bigcap_{k+1\le i\le n}N_{f_i}$ by Corollary 1 and $\alpha\not\in W_1$ but $\alpha\in W_2$, then there must be some $k+1\le i\le n$ with $\alpha\not\in N_{f_i}$, so $f_i(\alpha)\ne 0$ and therefore $f_i\in W_1^0$ but $f_i\not\in W_2^0$.