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?

  • $\begingroup$ Right, yes, I'll edit it! $\endgroup$ – dhk628 Nov 20 '15 at 4:50

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.

  • 1
    $\begingroup$ 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 $T$. $\endgroup$ – user160738 Nov 20 '15 at 3:10
  • $\begingroup$ Ah, I see I misread the question, let me edit! $\endgroup$ – Servaes Nov 20 '15 at 3:12
  • $\begingroup$ Thank you for the answer! Could you explain why $\phi$ factors over $V/\ker T$ and why that implies $\phi=\psi\circ\pi$? $\endgroup$ – dhk628 Nov 20 '15 at 5:02
  • $\begingroup$ 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. $\endgroup$ – Servaes Nov 20 '15 at 5:28

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.