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I am not sure if i'm understanding this correctly but if I have a map from $T:X \rightarrow Y$ then the adjoint $T^*: Y \rightarrow X$

Is the inverse not defined in the same way? I figure I am missing some assumption or restriction.

I would really appreciate someone clarifying this.

knk
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    $X$ and $Y$ are inner product spaces? ... Note that $T$ may certainly not be an isomorphism, so an inverse won't make sense at all. The adjoint is defined using an inner product (or else in terms of dual spaces). – Ted Shifrin Sep 11 '17 at 21:30
  • I have not delved into dual spaces. Thanks for pointing out that T may not be an isomorphism. However what would happen If we add the isomorphic constraint? – knk Sep 11 '17 at 21:38
  • In finite dimension, refering to canonical basis, you can identify operators and matrices: taking the adjoint can be identified with taking the transposed matrix (or the transpose conjugate in case we are in $\mathbb{C^n}$), while the inverse is identified (surprise surprise) with ... the inverse operation. Let us come to your question which, in matrix formulation, becomes : can we have $A^T=A^{-1}$ ? The answer is yes in the very special case of an orthogonal matrix. – Jean Marie Sep 11 '17 at 21:52

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The adjoint $T^*$ is defined by $\langle x,T^*(y)\rangle_X = \langle T(x),y\rangle_Y$ for all $x\in X$ and $y\in Y$.

To answer your follow-up question, you can check, for example, that for $X=Y=\Bbb R^n$, you'll have $T^* = T^{-1}$ if and only if the matrix of $T$ with respect to an orthonormal basis is an orthogonal matrix (i.e., the columns form an orthonormal basis for $\Bbb R^n$).

Ted Shifrin
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