image of adjoint equals orthogonal complement of kernel

Question

Let $T:V\to W$ be a linear map of finite-dimensional spaces. Then $${\rm im}(T^{\textstyle*})=({\rm ker}\,T)^\perp\ .\tag{$*$}$$ I can prove this as follows: $${\rm ker}(T^{\textstyle*})=({\rm im}\,T)^\perp\tag{$**$}$$ is quite easy, and we know $T^{\textstyle*}{}^{\textstyle*}=T$ and $W^{\perp\perp}=W$, so $${\rm im}(T^{\textstyle*})=({\rm im}\,T^{\textstyle*})^{\perp\perp}=({\rm ker}\,T^{\textstyle*}{}^{\textstyle*})^\perp=({\rm ker}\,T)^\perp\ .$$ However, I would be interested in a "direct" proof of $(*)$. It's fairly easy to show ${\rm LHS}\subseteq{\rm RHS}$. For the converse I have tried obvious things but seem to be going round in circles.

Also, any insights as to why $(**)$ is harder than $(*)$ - if in fact it is :)

Edit. To clarify, I am considering the adjoint defined in terms of an inner product, $$\langle\,T({\bf v})\mid{\bf w}\,\rangle =\langle\,{\bf v}\mid T^{\textstyle*}({\bf w})\,\rangle\ .$$

Answer 1

Let ${\bf v}\in{\rm im}(T^\ast)$, then ${\bf v}=T^\ast({\bf w})$ for some ${\bf w}\in W$. Now, given ${\bf u}\in\ker T$, we see that $T({\bf u})={\bf 0}$ and therefore $$\langle {\bf u}\mid{\bf v}\rangle =\langle {\bf u}\mid T^\ast({\bf w})\rangle =\langle T({\bf u})\mid {\bf w}\rangle =\langle {\bf 0}\mid{\bf w}\rangle =0.$$ That is, ${\bf v}\in(\ker T)^\perp$. Conversely, if ${\bf v}\notin{\rm im}(T^\ast)$, then there exists an ${\bf v}'\in{\rm im}(T^\ast)^\perp$ such that $\langle {\bf v}\mid{\bf v}'\rangle\ne 0$. In fact, we have ${\bf v}'\in\ker T$ because $T^\ast T({\bf v}')\in{\rm im}(T^\ast)$, which implies $$\langle T({\bf v}')\mid T({\bf v}')\rangle =\langle {\bf v}'\mid T^\ast T({\bf v}')\rangle=0\quad\Longrightarrow\quad T({\bf v}')={\bf 0}.$$ Therefore ${\bf v}\notin(\ker T)^\perp$, which completes the proof.

Answer 2

$\newcommand{\im}{\mathrm {im}}$ For the inclusion $\im(T^*)\subseteq (\ker T)^\perp$, let $\eta\in \im(T^*)\subseteq V^*$. Then there exists $\xi\in W^*$ such that $\xi\circ T = T^*(\xi) = \eta$. Hence, for every $x\in \ker T$, we have $$ \eta(x) = \xi\bigl(T(x)\bigr) = 0, $$ i. e. $\eta\in (\ker T)^\perp$.

Now, you can conclude by noting that $\dim \im(T^*) = \dim(\ker T)^\perp$. Alternatively, you can proceed as follows:

For the inclusion $(\ker T)^\perp\subseteq \im(T^*)$, let $\eta\in (\ker T)^\perp\subseteq V^*$. This means $\eta\colon V\rightarrow K$ is a $K$-linear map such that $\ker T\subseteq \ker \eta$. By the homomorphism theorem, this induces a $K$-linear map $\overline\eta\colon V/\ker T\rightarrow K$, also $T$ induces an isomorphism $\overline T\colon V/\ker T\rightarrow \im T$. This gives us a $K$-linear map $$ \xi'\colon \im T\xrightarrow{\overline T^{-1}} V/\ker T\xrightarrow{\overline\eta} K $$ such that $\xi'\circ T = \eta$.
Since we can complete a basis of $\im T$ to a basis of $W$, we find a $K$-linear map $\xi\colon W\rightarrow K$ such that $\xi\big|_{\im T} = \xi'$. Now, it follows that $$ T^*(\xi) = \xi\circ T = \xi'\circ T = \eta $$ and hence $\eta\in \im(T^*)$.

The reason why ($**$) is so much easier than ($*$) is that in ($**$) you don't have to construct linear maps. If you want to show $W = W^{\perp\perp}$ without using a dimension argument you would also have to construct linear maps.

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As a side remark: If you know exact sequences and that taking duals is exact (actually left-exactness suffices, which is a general property of the $\hom$-functor), then you can show $\im(T^*)=(\ker T)^\perp$ as follows: The exact sequence $$ 0\longrightarrow \ker T\longrightarrow V\xrightarrow T\im T\longrightarrow 0 $$ gives an exact sequence $$ 0\longrightarrow (\im T)^*\longrightarrow V^*\xrightarrow f (\ker T)^*\longrightarrow 0. $$ By definition, we have $(\ker T)^\perp = \ker f$ and exactness gives $\ker f = (\im T)^*$ (where we view $(\im T)^*\subseteq V^*$ via the injective map given by precomposing $T$). So it suffices to show $(\im T)^* = \im T^*$. But this follows directly from the definitions.
Here, all the work done above is implicit in the exactness of taking duals.