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How would you go about proving whether or not the adjoint of a linear operator $H$ (in a vector space) is also linear? I have

$$H: X \to Y \text{ and } H^*: X \to Y$$

such that

$$\langle H(x),y\rangle = \langle x,H^*(y)\rangle$$

I'm not sure where to start, any hints would be great and I can take it from there.

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    $\begingroup$ Note my edits: Proper usage is $X\to Y$, not $X->Y$, and $\langle H(x),y\rangle$, not $<H(x),y>$. $\qquad$ $\endgroup$ – Michael Hardy Sep 28 '16 at 2:41
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    $\begingroup$ Yes it is true. Show that $H^*(u + v) = H^*(u) + H^*(v)$ $\endgroup$ – IAmNoOne Sep 28 '16 at 2:43
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    $\begingroup$ Shouldn't it be $H^*: Y\to X$? $\endgroup$ – user137731 Sep 28 '16 at 2:43
  • $\begingroup$ I'm not sure. The problem i'm looking at says $H^*: X \to Y$ $\endgroup$ – Ghazal Sep 28 '16 at 2:45
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    $\begingroup$ It must be a typo, then, since $\langle H(x),y\rangle$ only makes sense if $y \in Y$ (since $H(x) \in Y$), and $\langle x,H^*(y)\rangle$ only makes sense if $H^*(y) \in x$ (since $x \in X$). $\endgroup$ – arkeet Sep 28 '16 at 2:47
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Hint for additivity: Show that $\langle x, H^*(y) + H^*(y') - H^*(y+y') \rangle = 0$ for all $x \in X$ and $y,y' \in Y$.

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