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I solved it like this : $$\langle (f^*)^*(v),w \rangle=\langle v,f^*(w)\rangle=\langle f(v),w\rangle$$ My lecture notes gave a proof with some more steps. Now i'm not sure, maybe i messed something.

here $f^*$ denotes the adjoint of the linear map $f$.

Mikasa
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sigmatau
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  • For linear maps you can attach them a matrix. You should try this proof using matrices. – Iuli Aug 11 '13 at 15:50
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    I guess the adjoint was defined as (for all $v,,w$) $\langle v,, f(w)\rangle = \langle f^\ast(v),, w\rangle$ and the proof in the notes went something like $\langle (f^\ast)^\ast(v),, w\rangle = \langle v, f^\ast(w)\rangle = \overline{\langle f^\ast(w), v\rangle} = \overline{\langle w, f(v)\rangle} = \langle f(v),w\rangle$? – Daniel Fischer Aug 11 '13 at 15:51
  • yes @DanielFischer that was the proof. And the definition we gave is as you stated but it starts in the first slot. Am i missing something in my resoning? – sigmatau Aug 11 '13 at 15:52
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    Well, if you have something that you may use for the equality $\langle f(x),y\rangle = \langle x, f^\ast(y)\rangle$, it is okay. That's a trivial thing to prove of course, by hermitian symmetry, but if that is never done, your argument uses a property that hasn't been proved. – Daniel Fischer Aug 11 '13 at 15:55
  • @DanielFischer this equation is how we defined the adjoint function , so we should start from there i think. – sigmatau Aug 11 '13 at 15:57
  • @AmireBendjeddou Was it defined $\langle f(x),y\rangle = \langle x,f^\ast(y)\rangle$, or $\langle v, f(w)\rangle = \langle f^\ast(v), w\rangle$? The definition gives you one direction of moving the map to the other side off the inner product adding/removing a $^\ast$. Your chain of equations moves into both directions. At some point, you need to show that that is okay. – Daniel Fischer Aug 11 '13 at 16:00
  • @DanielFischer I got your point now. It was defined like the first equation of your last comment. Well now it makes sense , the second equation is not given so i can't go back from the second to the first slot. So i have to stick with the proof of mu lecture notes. Thanks for clearing my doubt! – sigmatau Aug 11 '13 at 16:04

3 Answers3

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Once you have $\langle (f^*)^*(v),w \rangle=\langle f(v),w\rangle$ for all $v$ and $w$ you still need to do a little bit more to show $(f^*)^* = f$. First rewrite it as $\langle ((f^*)^* - f)(v),w \rangle=0$ which is true for all $v$ and $w$. In particular it is true for $w = ((f^*)^* - f)(v)$. So we get $\langle ((f^*)^* - f)(v),((f^*)^* - f)(v) \rangle=0 \implies ((f^*)^* - f)(v) = 0$ for all $v$. And so $(f^*)^* - f = 0 \implies (f^*)^* = f$.

Pratyush Sarkar
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From your argument, you can conclude that, for $y = (f^{\ast})^{\ast}(v) - f(v)$, one has $$ \langle y,w \rangle = 0 \quad \forall w $$ And you want to conclude that $y = 0$. For this you need to know that these linear functionals $$ \varphi_w : x \mapsto \langle x,w\rangle $$ separate points in your vector space. For this you will need the Riesz representation theorem and the Hilbert space version of the Hahn-Banach theorem. Is that what your book uses?

Prahlad Vaidyanathan
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I assume we are working in a finite dimensional inner product space V. What your proof is missing is the well known fact that if $f:V \to V$ and $g:V \to V$ are linear transformations with $<f(v),w>=<g(v),w>$ for all $v,w \in V$, then $f=g$.
To prove this fact, let $\{e_1, ..., e_n\}$ be an orthonormal basis of $V$ (you can construct one via the Gram-Schmidt process). Then for each $v \in V$ we have the identity $v= \sum_{i=1}^n<v,e_i>e_i$. In particular $$f(e_j) = \sum_{i=1}^n <f(e_j), e_i>e_i = \sum_{i=1}^n<g(e_j),e_i>e_i = g(e_j)$$ Thus, $f$ and $g$ agree on a basis, so they are equal.

walcher
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