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Alright, suppose we are given $V$, a finite dimensional inner product space, and a linear map, $T:V \rightarrow V$, with its corresponding adjoint, $T^\star :V \rightarrow V$. I want to show:

$[im(T)]^\perp = ker(T^\star)$.

However I just, for example, start with the left side and go as this:

$[im(T)]^\perp = \{ v \in V | \langle v,T(x) \rangle=0, \forall x \in V \}$

which I start working on as so:

$= \{ v \in V | \langle v,T(x) \rangle =0, \forall x \in V \}$

$= \{ v \in V | \langle T^\star (v),x \rangle =0, \forall x \in V \}$

But now I get stuck and I really just want to show $T^\star(v)=0$ in my set, but cannot manage to find the logic to get there. Any insight?

anak
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1 Answers1

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Isn't $x$ arbitrary? What can you infer from the properties of the (nondegenerate) inner product?

  • So you are suggesting I pick a special $x$ since it must hold true for all $x$s? – anak Dec 10 '14 at 01:26
  • @anakhronizein No, that $x$ need not be $0$, so $T^\star(v)$ must be... –  Dec 10 '14 at 01:28
  • I don't quite follow, but letting $x=T^\star (v)$ means that $\langle T^\star (v),T^\star (v) \rangle = 0$ implies $T^\star (v)=0$ by the iff on the 4th axiom on that page... – anak Dec 10 '14 at 01:35
  • @anakhronizein Look at the second paragraph. You're assuming the inner product is nondegenerate, so $T^\star(v)=0$ in your set. –  Dec 10 '14 at 01:41
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    @anakhronizein Here's a simple analogy. If $x\times y=0$, and $y$ is any arbitrary integer/real number/complex number/..., what must $x$ be? –  Dec 10 '14 at 01:42
  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Mark Fantini Dec 10 '14 at 01:50
  • @Sanath Oh. I see now. Thank you. – anak Dec 10 '14 at 01:56
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    @MarkFantini This is a hint, not a clarification (but I understand why it can be misunderstood like a clarification). –  Dec 10 '14 at 01:58