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For a linear operator it holds that $\ker (T^\ast ) = (\operatorname{ran} (T))^\perp$. The star denote the adjoint of $T$ and $\perp$ the orthogonal complement. Is there a geometric intuition for the meaning of $\ker (T^\ast ) = (\operatorname{ran} (T))^\perp$?

blue
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  • Here's a recent related thread. – littleO Aug 23 '13 at 08:01
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    For what it's worth, there's a quick proof: $x \in N(T^) \iff T^(x) = 0 \iff \langle T^*(x),y \rangle = 0 \forall y \iff \langle x, T(y) \rangle = 0 \forall y \iff x \in R(T)^{\perp}$. – littleO Aug 23 '13 at 08:07
  • @littleO Thank you I did not know there is a theorem called the fundamental theorem of linear algebra! – blue Aug 23 '13 at 09:21
  • Sure. In that thread, there are some comments questioning whether the theorem should really be called the "fundamental theorem" of linear algebra (although I do think the theorem should be emphasized more in undergrad linear algebra classes, regardless of what it's called). There are also some interesting comments about a similar but more general theorem that doesn't use inner product spaces. – littleO Aug 23 '13 at 09:37

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There is a related question on an intuitive view on adjoints at mathoverflow. Although not directly related to $\ker T^*=(\operatorname{ran}T)^\perp$, it may help your geometric intuition:

Considering the graph of $T:\mathcal{H}\to\mathcal{H}$, $\mathcal{G}(T):=\{(x,Tx)\mid x\in\mathcal{H}\}$ we can naturally equip the product space $\mathcal{H}\times\mathcal{H}$ with an inner product by setting $\langle(x,y),(x',y')\rangle:=\langle x,x'\rangle+\langle y,y'\rangle$ and therefore look at the orthogonal complement of $\mathcal{G}(T)\subset\mathcal{H}\times\mathcal{H}$ in the ambient product space $(y,z)\in\mathcal{G}(T)^\perp$: $$0=\langle(x,Tx),(y,z)\rangle=\langle x,y\rangle+\langle Tx,z\rangle=\langle x,y\rangle+\langle x,T^*z\rangle\qquad\Rightarrow\qquad y=-T^*z$$ Thus, the adjoint $T^*$ is the operator such that $\mathcal{G}(T^*)=-(\mathcal{G}(T))^\perp$, i.e. the operator which (up to an sign) has the graph orthogonal to $T$.

Community
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gofvonx
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For the sake of simplicity assume we have an inner product so the dual space of each vector space is naturally isomorphic to the vector space.

Let $T: V \to W$ be a homomorphism between vector spaces and $T' : W' \to V'$ be the adjoint. Now the image of adjoint $\mathrm{Im}\;T'$ is all non-negligible elements of the domain $\mathrm{Dom}\;T$ of $T$.

If none-of the elements of $\mathrm{dom}\;T$. are negligible, then $$ \mathrm{Im}\;T' = \mathrm{Dom}\;T $$

In fact $T'$ is surjective whenever $T$ is injective and vice versa. In other words, the image of $T'$ is $V$ as seen from $W$ through the lens of $T$.

LMZ
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