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Let $V=M^{C}_{n \times n}$ with the standard inner product.
Suppose $T_{P}$ is a linear transformation from $V$ to $V$ such as: $\forall X \in V,\ T_{P}X=P^{-1}XP$ for some invertible matrix $P$.

Show that the following holds: $(T_{P})^{*}=T_{P^{*}}$ (that is, the conjugate transpose of T equals to T with P conjugate transposed).

In practice I need to show that $(T^{*}_{P})X=(P^*)^{-1}XP^*$, since $(P^{-1})^{*}=(P^{*})^{-1}$.

The examples I've seen so far showed that to find the conjugate transpose of a linear transformation, one needs to find the transformation matrix with respect to some orthonormal basis, and then use the relation $[T^*]_{B}=[T]_{B}^*$. But I somehow doubt that that needs to be done here.

  • Are there any properties of the conjugate transpose of a linear transformation that can help me here?
  • The linear transformation itself reminds me of matrix similarity, does it have any meaning?
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The inner product in $V$ is $(X,Y)=Tr(X^*Y)$. You therefore need to prove that $Tr(X^*P^{-1}YP)=Tr(({P^*}^{-1}XP^*)^*Y)$. I hope I didn't reveal too much. – user8268 Mar 25 '11 at 13:02
@user8268: You have the Frobenius product backwards. $\langle X,Y\rangle = \mathrm{trace}(Y^*X)$, not $\mathrm{trace}(X^*Y)$. – Arturo Magidin Mar 25 '11 at 13:42
@AM: it's a matter of convention and certainly doesn't change anything in this problem :) – user8268 Mar 25 '11 at 13:46
up vote 1 down vote accepted

There are a couple of things you can do, depending on how you defined the adjoint of a linear operator.

  1. If you defined $(T_P)^*$ as the unique linear transformation with the property that for all $A$ and $B$, $\langle T_P(A),B\rangle = \langle A,(T_P)^*(B)\rangle$, where $\langle\cdot,\cdot\rangle$ is the inner product of the space (which in this case is given by $\langle X,Y\rangle = \mathrm{Trace}(Y^*X)$, the Frobenius inner product, with $Y^*$ being the conjugate transpose). Then you need to show that $T_{P^*}$ has the "right" property. That is, that if $A$ and $B$ are matrices, then $$\langle T_P(A),\rangle B = \langle A , T_{P^*}(B)\rangle.$$ So compute: $$\begin{align*} \langle T_P(A),B\rangle &= \langle P^{-1}AP,B\rangle = \mathrm{Trace}(B^*(P^{-1}AP)).\\ \langle A,T_{P^*}(B)\rangle &= \langle A,(P*)^{-1}BP\rangle = \mathrm{Trace}((P^*)^{-1}BP^*)^*A). \end{align*}$$ Now use properties of the conjugate transpose and the trace to see if they are equal.

  2. If you defined $(T_P)^*$ in terms of the matrix relative to an orthonormal basis, then take the standard orthonormal basis of $M_{n\times n}$, which consists of the matrices $E_{ij}$ that have a $1$ in the $(ij)$th entry and $0$s elsewhere, and see what $P^{-1}E_{ij}P$ is.

The fact that it reminds you of matrix similarity is just the consequence that matrix similarity is "really" defined in terms of automorphisms: conjugating by an invertible matrix is an automorphism of the $n\times n$ matrices, and similarity consists of the orbits of the action of the group of these automorphisms.

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Thanks for the detailed explanation. Given a certain matrix $P$, to find the transformation basis of $(T_{P})*$ in the standard basis, I need to simply find the transformation basis of $T_{K}$ where $K=P^*$, right? (based on the above) – daniel.jackson Mar 25 '11 at 14:26
Or simply find the transformation basis of $T_{P}$ and take its transpose-conjugate, which should yield the same result. – daniel.jackson Mar 25 '11 at 14:58
I don't know what you mean by "transformation basis" of a linear transformation... To find the coordinate matrix of a linear transformation, you look at the images of the basis. – Arturo Magidin Mar 25 '11 at 15:04
Yes that's what I meant.. Or transformation matrix. Thanks. – daniel.jackson Mar 25 '11 at 15:09

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