I'm given a space of $X \in R^{mxn}$ with inner product $\langle X_1, X_2 \rangle = tr(X_1^T X_2)$ and another space of random vectors $Y \in RV^m$ with inner product $\langle y_1,y_2 \rangle = E[y_1^Ty_2]$. Additionally an operator that maps between these spaces $A: X \rightarrow Y $ is known to be $A(X) = Xb(\omega)$. Where $b \in RV^n$ and $E[bb^T] = I$. I believe the last identity is only necessary in proving that $A(X)$ is one to one.

The goal is to solve the least squares problem for the case where A is one to one. Therefore I know it's necessary to determine the adjoint operator, $A^* : Y \rightarrow X$, which satisfies the equation $$ \langle y,A(X)\rangle = \langle A^*(y),X \rangle$$

But with different definitions for the inner products of the two spaces, I am stuck in manipulating the expectation into a function with the trace. I have...

$$ \langle y,A(X)\rangle$$ $$E[y^TXb(\omega)]$$ $$.$$ $$.$$ $$.$$ $$tr((something)^TX)$$ $$\langle A^*(y),X \rangle$$

If anyone has any references on dealing with such a combination of different inner products in determining the adjoint operator, especially in a stochastic framework, I would be very nice to see. General literature with examples of adjoint determination would be greatly appreciated as well.


1 Answer 1


Using the cyclic property of trace and linearity of expectation/trace we have: \begin{eqnarray} \langle y, A X \rangle &=& E [ y^T X b ] \\ &=& E [ \operatorname{tr} ( y^T X b )] \\ &=& E [ \operatorname{tr} ( b y^T X )] \\ &=& \operatorname{tr} ( E [ b y^T ] X ) \\ &=& \operatorname{tr} (E[y b^T]^T X ) \\ &=& \langle E[y b^T] , X \rangle \end{eqnarray} Hence $A^*y = E[y b^T]$.

  • $\begingroup$ Of course... because $y^TXb$ is scalar valued, and $X$ is itself not a random vector/matrix. It just had to be something this obvious. :) $\endgroup$
    – Falimond
    Nov 8, 2015 at 5:22
  • $\begingroup$ Well, I would say that it is elementary but not obvious :-). Many things become obvious in retrospect. $\endgroup$
    – copper.hat
    Nov 8, 2015 at 5:26
  • $\begingroup$ Yes, hindsight is 20/20. Oh how convenient that inner products be scalar valued by definition! $\endgroup$
    – Falimond
    Nov 8, 2015 at 5:39

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