Let $M_n$ be space of compex n x n matrices with inner product $(A,B)= Tr A\bar{B}^t$. Find adjoint operators:

i. For operator of left multiplication $L_A :X\rightarrow AX$ by matrix A for $X \in M_n$,

ii.For operator of right multiplication $R_A:X\rightarrow XA$ by matrix A for $X \in M_n$,

iii. and for similarity operator $C_T:X\rightarrow T^{-1}XT$ for invertible matrix T.

  • $\begingroup$ I did not have any clear idea. @ThibautDumont $\endgroup$ – abcdef Apr 17 '15 at 19:27

For every matrix $A$, I use the notation $A^*=\bar{A}^t$. Also if $T$ is a linear operator $T:M_n\rightarrow M_n$, I denote by $T^*$ the adjoint of $T$.

Now the adjoint operator $T^*$ associated to $T$ is the only operator satisfying $$ (TX,Y)=(X,T^*Y)$$ for all $X,Y\in M_n$. In the first case, we need to find $L_A^*$ such that
$$ Tr(AXY^*)=Tr(X{L_A}^*(Y)).$$

Try to use that for every pair $X,Y\in M_n$, we have $(XY)^*=Y^*X^*$ and $Tr(XY)=Tr(YX)$ to conclude.


Here it is for the first case:

The property of the trace implies $$ Tr(AXY^*)=Tr(XY^*A).$$ But since $(A^*)^*=A$, it follows that $$Tr(XY^*A)=Tr(X(A^*Y)^*).$$ Altogether we obtained $$(AX,Y)=(X,A^*Y).$$ We see that the adjoint of $L_A$ is the multiplication on the left by $A^*$. In other words, $(L_A)^*=L_{A^*}$.

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  • $\begingroup$ Could you be more explicit please. What you have already mentioned above we have taken as definitions in class @ThibautDumont $\endgroup$ – abcdef Apr 17 '15 at 20:12
  • $\begingroup$ @DI... I let you figure out the remaining cases. $\endgroup$ – Thibaut Dumont Apr 18 '15 at 8:07
  • $\begingroup$ Thank you! @ThibautDumont I got it. $\endgroup$ – abcdef Apr 18 '15 at 16:32
  • $\begingroup$ Can we write a (XA,Y)=(X,YA*)?? @ThibautDumont $\endgroup$ – abcdef Apr 18 '15 at 19:36
  • $\begingroup$ I still do not believe I have been able to approach this question. Could you please help me? @ThibautDumont $\endgroup$ – abcdef Apr 21 '15 at 13:06

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