Today I randomly encountered this linear algebra question:
Let $V$ be a finite dimensional vector space. If $f_1,f_2,\ldots,f_n,g\in V^\vee$, then $g$ is in the span of ${f_i}$ if and only if $\bigcap_{i=1}^n\ker f_i\subset\ker g$.
This is of course easy (and important?). Note that each of the $\ker f_i$ is a hyperplane in $V$. Also WLOG we may assume the $f_i$’s are linearly independant by reductio with the trivial $(\Longrightarrow)$ direction. But the dimension of functionals vanishing on $\bigcap_{i=1}^n\ker f_i$ is at most $n$,
\[n=n.\qquad\square\]The question proceeds with
What if $V$ is infinite dimensional?
Before any of that, perhaps we should rephrase the statement as follows.
Proposition. Every functional that annihilates $\bigcap_{i=1}^n\ker f_i$ is a linear combination of $f_i$’s.
And further? Since $f_i$ is uniquely determined by $\ker f_i$ up to a scalar…
Proposition’. For subspaces $U,W$ of a finite-dimensional $V$,
\[(U\cap W)^\bot=U^\bot+W^\bot.\]This. (Not quite.) At this point we can still use the previous dimension argument or even taking bases by hand to prove this, but that’s kind of pointless. Instead we can keep rewriting it until the problem cease to exist.
Recall that on fields everything is injective and $(-)^\vee$ is exact, and we have the most desirable dual properties:
\[0\longrightarrow U\longrightarrow V\longrightarrow V/U\longrightarrow 0\] \[0\longrightarrow U^\bot=(V/U)^\vee\longrightarrow V^\vee\longrightarrow U^\vee\longrightarrow 0\] \[0\longrightarrow U^{\vee\vee}=U^{\bot\bot}\longrightarrow V^{\vee\vee}\longrightarrow U^{\bot\vee}\longrightarrow 0\]We see that after identifying spaces with double-dual we can directly compute
\[\text{LHS}\simeq(U^{\bot\bot}\cap W^{\bot\bot})^\bot=(U^\bot+W^\bot)^{\bot\bot}\simeq\text{RHS}.\]
Here we have an issue. Double-dual no longer works if $\dim U$ or $\dim W$ or $\dim(U^\bot+W^\bot)$ is not finite. Nevertheless, by the first duality, we still have
Lemma. $(\ker f)^\bot$ is spanned by $f$.
Proof: $\dim(\ker f)^\bot=\dim(\operatorname{im}f)^\vee=1$.
And we can generalize Proposition’ by hand.
Proposition’1. If $U, W$ are subspaces of a vector space $V$, and $\dim(W^\bot)<\infty$, then $(U\cap W)^\bot=U^\bot+W^\bot$.
First,
\[\dim\frac{V}{W}\leq\dim\left(\frac{V}{W}\right)^\vee=\dim W^\bot<\infty.\]In $U$ we have
\[\frac{U}{U\cap W}\simeq\frac{U+W}{W},\]thus
\[\dim\frac{U}{U\cap W}=\dim\frac{U+W}{W}\leq\dim\frac{V}{W}<\infty.\]Therefore, we can take a basis $(e_1,e_2,\ldots,e_k)$ of this space, with a corresponding dual basis $(\check{e}_1,\check{e}_2,\ldots,\check{e}_k)$, which, in view of the decomposition
\[V=(U\cap W)\oplus\frac{U}{U\cap W}\oplus\frac{V}{U},\]can be extended (by $0$) to elements of $V^\vee$. It is obvious now that if $f(U\cap W)=0$, then
\[\begin{align*} f&=\left(f-\sum_{i=1}^kf(e_i)\check{e}_i\right)+\left(\sum_{i=1}^kf(e_i)\check{e}_i\right)\\ &\in U^\bot+W^\bot. \end{align*}\]
In a word: We see that doesn’t matter if $\dim V=\infty$. The codimensions are finite, and they behave nicely under (finite) intersections. We have the option to either use dimension argument (the one at the very beginning of this article) or duals.
Post date: 2024/11/04
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