Hoffman and Kunze, Linear Algebra Chapter 3 theorem 20

Question

Theorem $20.$ Let $g,f_1,\dots ,f_r$ be linear functionals on a vector space $V$ with respective null spaces $N,N_1,\dots ,N_r.$ Then $g$ is a linear combination of $f_1,\dots ,f_r$ if and only if $N$ contains the intersection $N_1\cap \cdots \cap N_r.$

I am given a lemma for this theorem:

If $f$ and $g$ are linear functionals on a vector space $V,$ then $g$ is a scalar multiple of $f$ if and only if the null space of $g$ contains the null space of $f.$

Attempt : Forward implication is trivial. I am trying to prove the reverse implication by induction.

$r=1$ is handled by the lemma. Suppose the statement is true for $r=k-1$ i.e., $$\bigcap_{i=1}^{k-1} N_i\subseteq N \implies g=\sum_{i=1}^{k-1}c_if_i.$$

Also we are given that $\bigcap_{i=1}^{k} N_i \subseteq N.$ We have to show $g=\sum_{i=1}^{k}c_if_i.$

By sketching out some venn diagrams I noticed that $\bigcap_{i=1}^{k} N_i \subseteq N$ does not necessarily mean $\bigcap_{i=1}^{k-1} N_i \subseteq N.$

But if I consider the restriction functions $g',f_1^{'},\dots ,f_{k-1}^{'}$ on the subspace $N_k$ then I have $\bigcap_{i=1}^{k-1} N_i^{'} \subseteq N$ so that $g'=\sum_{i=1}^{k-1}f_i^{'}.$

What next?

These are related posts :$(1)$,$(2)$

Answer 1

Here is a more down-to-earth proof via induction:

Assume $N_1 \cap \dotsc \cap N_r \subset N$

Note that $\dim (N_1 \cap \dotsc \cap N_{r-1}) \leq \dim (N_1 \cap \dotsc \cap N_r) +1$, i.e. the dimension can only drop by at most one, since you intersect with a hyperplane.

This shows that there is some vector $v$, s.t. $N_1 \cap \dotsc \cap N_{r-1} \subset N_1 \cap \dotsc \cap N_r + \langle v \rangle$.

If $f_r(v)=0$, we actually have $N_1 \cap \dotsc \cap N_{r-1} = N_1 \cap \dotsc \cap N_r$ and there is nothing to show, because we can immediately do induction. Otherwise we can rescale and assume $f_r(v)=1$.

Now consider $h := g-g(v)f_r$. Note that $h(v) = g(v)-g(v)=0$.

Thus we have $h(N_1 \cap \dotsc \cap N_{r-1}) \subset h(N_1 \cap \dotsc \cap N_{r}) + \langle h(v) \rangle = h(N_1 \cap \dotsc \cap N_{r})$

If $x \in N_1 \cap \dotsc \cap N_{r}$, then $x \in N$ and $g(x)=0$. But also $x \in N_r$ and $f_r(x)=0$. This shows $h(x)=0$ and we have finally shown that the nullspace of $h$ contains $N_1 \cap \dotsc \cap N_{r-1}$. Induction yields that $h = g-g(v)f_r$ is a linear combination of $f_1, \dotsc, f_{r-1}$, q.e.d.

Answer 2

I would view this as a formal consequence of a very basic property of the functor $V \mapsto V^*$, namely $(\operatorname{ker} \varphi^*)^* = \operatorname{im} \varphi$.

Consider the map $$K^r \to V^*, e_i \mapsto f_i.$$ Dualizing yields a map $$\varphi^*: V \to (K^r)^*, v \mapsto \sum_{i=1}^r f_i(v)e_i^*.$$

You are asking whether $g \in \operatorname{im}\varphi = (\operatorname{ker} \varphi^*)^* =(N_1 \cap \dotsc N_r)^*$.

By definition this is equivalent to $g(N_1 \cap \dotsc N_r)=0$, i.e. $\operatorname{ker} g = N$ contains $N_1 \cap \dotsc N_r$.