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Suppose $T:V\to W$ and that $V$ is finite-dimensional.

I want to prove that $$\text{Im }T'=(\ker T)^0$$ where $T'$ is the dual/transpose map and $(\ker T)^0$ is the annihilator of the kernel.

I know that $\phi \in V'$ is an annihilator of $\ker T$ if and only if $$\phi(v)=0 \space \forall v\in \ker T$$ if and only if $$\phi(v)=0 \space \forall v \in V \text{ such that } T(v)=0$$

Now I also know that $\phi \in \text{Im }T'\subset V'$ if and only if $$\phi = f \circ T \text{ for some } f \in W'$$ I can see that if $v\in \ker T$, then this implies $\phi(v) = f(T(v)) = 0$, so $\phi \in (\ker T)^0.$

However, I'm having trouble showing the other way, that if $\phi \in (\ker T)^0$, then $\phi \in \text{Im }T'.$

Could anybody help?

Servaes
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quasicoherent_drunk
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2 Answers2

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By the first isomorphism theorem the map $T$ factors over $V/\ker T$ as $$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\overline{T}}{\longrightarrow}\ \operatorname{im}T,$$ where $\pi$ is the canonical quotient map and $\overline{T}$ is an isomorphism. Note that $\pi=\overline{T}^{-1}\circ T$.

If $\phi\in(\ker T)^{\circ}$ then $\ker T\subseteq\ker\phi$. This means $\phi$ also factors over $V/\ker T$, as $$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\psi}{\longrightarrow}\ F,$$ where $F$ denotes the base field, and $\psi\in(V/\ker T)'$. It follows that $$\phi=\psi\circ\pi=\psi\circ(\overline{T}^{-1}\circ T)=(\psi\circ\overline{T}^{-1})\circ T,$$ where of course $\overline{T}^{-1}$ is an isomorphism, hence $\psi\circ\overline{T}^{-1}\in(\operatorname{im}T)'$. Because $\operatorname{im}T$ is a linear subspace of $W$, the linear functional $\psi\circ\overline{T}^{-1}$ on $\operatorname{im}T$ extends to a linear functional $f$ on $W$, which then satisfies $$\phi=f\circ T,$$ by construction, as desired. Note that $f$ is far from unique in general.

Servaes
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    I'm rather confused by what you are doing here. How is it that image of $\phi$ is a subspace of $W$, when $\phi$ is supposed to be a linear functional? Also, I cannot see why this $f$ must be linear on "whole" of $W$, not just in image of $T$. – user160738 Nov 20 '15 at 03:10
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    Ah, I see I misread the question, let me edit! – Servaes Nov 20 '15 at 03:12
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    Thank you for the answer! Could you explain why $\phi$ factors over $V/\ker T$ and why that implies $\phi=\psi\circ\pi$? – quasicoherent_drunk Nov 20 '15 at 05:02
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    It factors over $V/\ker T$ because it factors over $V/\ker\phi$ and $\ker T\subseteq\ker \phi$. You could also try writing out what the maps $\pi$ and $\psi$ look like. – Servaes Nov 20 '15 at 05:28
  • Where did you use finite dimension in the proof? It is wrong when finite dimension, am I right? – Hyacinth Jan 16 '20 at 03:30
  • @Hyacinth In extending $\psi\circ\overline{T}^{-1}$ to $W$. – Servaes Jan 16 '20 at 21:44
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Because you already have that $(\ker T)^\circ\subseteq \text{im } T^*$ you can simply show that their dimensions must be equal and thus we know that they must be equal. We know that $$\dim V^*=\dim V=\dim\; \ker T+\dim\; \text{im }T$$ and that $$\dim V^*=\dim\; \text{im } T^*+\dim\; (\text{im } T^*)^\circ.$$ It is easy to show that $\text{rank } T^*=\text{rank }T.$ Thus we can re-write the equality and it follows immediately that $$\dim \ker T=\dim \; (\text{im } T^*)^\circ.$$ Therefore, the two must be equal.

user502145
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