1

Let $V$ be a vector space and $f$ a nonzero linear functional on $V$. Then $\text{dim image}(f)=1$ and $\text{dim null}(f)+\text{dim image}(f)=\text{dim}V$. And if $\text{dim}V<\infty$, then $\text{dim null}(f)=\text{dim}V-1$.

I'm confused as to why $V$ needs to be finite dimensional in order to conclude $\text{dim null}(f)=\text{dim}V-1$. Can't we just say that if $\text{dim}V\not<\infty$, we have $\text{dim}V-1=\text{dim}V$?

Stella Biderman
  • 31,475
user124910
  • 3,241
  • It's not an exact duplicate, but it follows as a special case from there which is good enough – Stella Biderman Jan 03 '17 at 11:18
  • @StellaBiderman If it isn't too much trouble, could you elaborate a bit further? To be clear, I'm not asking if the rank-nullity holds for infinite dimensional $V$, which seems to be what most of the discussions in that link concern. – user124910 Jan 03 '17 at 11:27

1 Answers1

2

You don't need $\dim V<\infty$ for this to hold, although it is often excluded because defining subtraction on cardinals is generally not done (though in this context it can be).

Here we have a discussion of the fact that the rank-nullity theorem holds for infinite dimensional $V$ using cardinal arithmetic. In this problem, we have the equation $1+\alpha=\alpha$ for $\alpha = \dim V$. Subtraction of a finite cardinal from an infinite cardinal can be defined and gives $\kappa - x = \kappa$ for all $\kappa\geq\aleph_0$ and $x<\aleph_0$. Therefore we have $\dim V - 1=\dim V=\dim\text{null}(f)$ just as you say.

The reason that defining subtraction for cardinals is often not done is that $\alpha-\beta$ is problematic when both $\alpha,\beta\geq\aleph_0$. If $\beta >\alpha$, then there isn't even an answer, since there aren't negative cardinals!

Community
  • 1
Stella Biderman
  • 31,475
  • 1
    And the conclusion is: one can certainly say that $\dim\operatorname{null}(f)=\dim V-1$, but it has no useful consequence at all unless $V$ is finite dimensional. – egreg Jan 03 '17 at 11:52