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Where $ f: V \rightarrow V $ is a linear operator on an inner product space $V$, I'm trying to show that $$ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $$

i.e. the Kernel of the adjoint is equal to the orthogonal complement of the image of f.

I taking the inner product of $f(v)$, where $v \in V$ and an arbitary element of the kernel, $k \in \mathrm{ker}(f^*) $

$$\langle f(v), k\rangle = \langle v, f^*(k)\rangle = \langle v, 0\rangle = 0$$

And then taking the inner product of $f(v)$ with $w$, where $w \in \mathrm{im}(f)^{\perp}$

By definition of the orthogonal complement to the image of f,

$$ \langle f(v), w\rangle = 0 $$

My question is, now that we have

$$ \langle f(v), k\rangle = \langle f(v), w\rangle = 0 $$

What can we conclude? I would like to conclude that because $w$ and $k$ are arbitary elements of their respective subspaces, that these must be equal, i.e. $ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $

However, I am not sure this is mathematically sound, is that conclusion valid or are there any technical barriers or additional statements that would have to be made to make that watertight?

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  • $\begingroup$ Observe that $\langle f(v),k\rangle=0$ holds for all $v\in V$, thus $\langle u,k\rangle=0$ for all $u\in \mathrm{im}(f)$, which implies $k\in (\mathrm{im}(f^*)^\perp$. $\endgroup$ – daw Oct 9 '14 at 18:55
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Let $v,w\in V$. Then $$ \langle f(v),w\rangle = \langle v, f^*(w)\rangle. $$ If $w\in \ker f^*$, then $\langle f(v), w\rangle =0$ for all $v\in V$, hence $w\in (\mathrm{im} f)^\perp$.

If $w\in (\mathrm{im} f)^\perp$ then $\langle v, f^*(w)\rangle=0$ for all $v\in V$, implying $f^*(w)=0$, hence $w\in \ker f^*$.

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