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Let $T \in L(V,W)$ where $L(V,W)$ denotes the set of linear maps from $V$ to $W$. Prove that $ \text {(null}~T^*)^\perp \subseteq \text{range}~T $ where $T^*$ is the adjoint operator ( not related to the adjoint matrix) and $A^\perp $ refers to the orthogonal complement of $A$. Reference source is linear algebra done right by sheldon axler

Attempt: Let $w_2 \in \text{(null}~T^*)^\perp$.

Our aim is to show that $\exists v \in V$ such that $Tv=w_2$.

We can express $W = \text {null} ~T ~\oplus~\text{(null}~T^*)^\perp$. Thus, Let $w = w_1+w_2$ where $w_1 \in \text {null} ~T$.

$T^*(w)=T^*(w_2)=v$ for some $v \in V$

If we somehow prove that $<(Tv-w_2),~(Tv-w_2)>=0$, then we are done

But expanding the above seems to make it only more complicated.

Any ideas on how to move ahead? I have browsed through similar questions and tried to employ similar techniques but in vain.

Thanks a lot!

MathMan
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  • @ArnaudD. That's a different question though. I tried to prove this question using similar techniques but in vain. – MathMan Dec 04 '19 at 12:26
  • It's the same if you swap $T$ and $T^*$, and if you take orthogonal complements on your inclusion. Also, there are a few other instances linked to that question, that could help as well. – Arnaud D. Dec 04 '19 at 12:26
  • @ArnaudD. The question that you refer to specifically asks about finite-dim spaces. – Arnaud Mortier Dec 04 '19 at 12:31
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    The result is false in infinite dimensional spaces. It is correct if you take the closure on RHS. – Kavi Rama Murthy Dec 04 '19 at 12:35
  • @KaboMurphy It can be false if one considers topological vector spaces and continuous linear maps, but without continuity considerations it also holds true in infinite-dimensional spaces. [Assuming choice] – Daniel Fischer Dec 04 '19 at 13:46
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    Does Axler accept bases, or should everything be done basis-free? (If the latter, I don't know how to prove it, and I think the assertion need not hold without choice.) – Daniel Fischer Dec 04 '19 at 13:48
  • @DanielFischer yes, basis is accepted. – MathMan Dec 04 '19 at 13:51
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    I'm not sure whether this works as a duplicate, because not every space is a dual. The ones assuming finite-dimensionality don't really work as duplicates. – Daniel Fischer Dec 04 '19 at 14:05
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    So, sketching the proof using bases: Let $w \notin T(V)$. Take a basis of $T(V)$, add $w$, and extend to a basis of $W$. The linear functional $\lambda$ mapping $w$ to $1$ and all other basis vectors to $0$ belongs to $\operatorname{null} T^{\ast}$, so $\lambda(w) \neq 0$ shows $w \notin (\operatorname{null} T^{\ast})^{\perp}$. Thus we have shown $W\setminus T(V) \subset W\setminus (\operatorname{null} T^{\ast})^{\perp}$, qed. – Daniel Fischer Dec 04 '19 at 14:10
  • @DanielFischer Thank you. great answer. I get the gist of it. i am trying to wrap my head around one point. I hope I can get back in case I face something, but thanks a lot! – MathMan Dec 04 '19 at 14:15

1 Answers1

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Assuming they are finite dimension vector spaces, we have that $rank(T^*) = rank(T)$. Therefore $\frac{W}{null(T^*)} \cong (nullT^*)^\perp \cong imT^* \cong imT$. Since $imT \subset (nullT^*)^\perp$ we have an equality by the dimensions.

HelloDarkness
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