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Given matrices P and Q defined as

P = {{x - I y, z}, {z, x + I y}}; Q = {{0, 1}, {1, 0}};

It is clear that $[P,QR]=0$ where $R$ is the complex-conjugation operation. My question is how can one find then the simultaneous eigenvectors of $P$ and $QR$?

EDIT-1: Thanks to the responses, I am updating the question with more details.

Why $PQR-QRP=0$ implies $QP-P^*Q=0$? This can be seen by letting $PQR-QRP$ to act on an eigenvector $\vec{x}$ of $P$ with eigenvalue $\lambda$, that is, $P \vec{x} = \lambda \vec{x}$. Noting that $R \vec{x}=\vec{x}^*$ (* is complex-conjugate and not conjugate-transpose), we have \begin{align} &(PQR)\vec{x} - (QRP)\vec{x}=0 \vec{x},\nonumber \\ \implies& (PQ) \vec{x}^* - (QR) \lambda \vec{x}=0,\nonumber \\ \implies& (PQ) \vec{x}^* - Q \lambda^* \vec{x}^*=0, \nonumber \\ \implies& (PQ) \vec{x}^* - Q P^* \vec{x}^*=0 \nonumber \\ \implies& PQ - Q P^* =0. \end{align}

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  • $\begingroup$ The trivial approach is to find all the eigenvectors of P and then see if they are eigenvectors of QR. Are you interested in matrices that are too large to make this practical? $\endgroup$
    – mikado
    Mar 19 at 16:38
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    $\begingroup$ Try Eigensystem[{P, Q}]. Probably you should convert the matrices first to real $4 x 4$ metrices... $\endgroup$
    – Henrik Schumacher
    Mar 19 at 17:20
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    $\begingroup$ @march, note that $[P, QR]=0$ implies $QP=P^*Q$ where $P^*$ is the complex conjugate of $P$. $\endgroup$
    – Mike
    Mar 19 at 19:38
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    $\begingroup$ I'm confused: complex conjugation does not seem to be a complex-linear transformation. Is $QR.v$ the same operation as Q. Conjugate[{1 + I, 2 + 2 I}] for $v=(1+i,2+2i)$? $\endgroup$
    – Michael E2
    Mar 19 at 20:06
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    $\begingroup$ Or to put it another way, how do you find the eigenvectors of $QR$? (In Mathematica code.) $\endgroup$
    – Michael E2
    Mar 19 at 20:17

1 Answer 1

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From the question and comments, the notation in the question means $$ [P, QR] = Q P - P^* Q = 0 $$ Indeed,

P = {{x - I y, z}, {z, x + I y}};
Q = {{0, 1}, {1, 0}};
FullSimplify[Q.P - ConjugateTranspose[P].Q, Element[{x,y,z}, Reals]]

returns

{{0,0},{0,0}}

Yet,

P.Q - Q.P

is not zero

{{0, -2 I y}, {2 I y, 0}}

Because $Q$ and $P$ do not commute they are not simultaneously diagonalizable, that is they do not have common eigenvectors.

However, $P$ and $P^*$ are indeed similar, so there exists a nonsingular $T$ such that $$ T^{-1} P T = P^* $$ and $$ Q P - T^{-1} P T Q = 0 \quad \implies \quad R P - P R = 0, \quad R = T Q $$ It is $R = T Q$ that share common eigenvectors with $P$. Is that what you were looking for?

Edit 1: After your response, what you are looking for cannot be described as a linear complex operator, as pointed out by @Michael E2 in the comments. It can however be described by a linear operator in the reals. In that case, you will be lifting $P$, $Q$, and $R$ into 4x4 real matrices, say $P_r$, $Q_r$, $R_r$, which you can do and will lead to $$ P_r Q_r R_r - Q_r R_r P_r = 0 $$ as you pointed out. Here $P_r$ and $Q_r R_r$ will be simultaneously diagonalizable by a matrix containing the four eigenvectors of $P_r$. And those will be just the two complex eigenvectors of the original matrix $P$. There will be two repeats that are linearly independent in the reals but not in the complexes.

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    $\begingroup$ I guess $[\cdot,\cdot]$ is not the standard $[A,B]=AB-BA$ commutator? If it is, why is $[P,QR]=PQR-QRP = QP-P^*Q$? Is there a (conjugate) transpose in there somewhere, and is the fact that $Q$ is symmetric and real being used? (Note the OP said $P^*$ was the conjugate, not the conjugate transpose, of $P$. But your derivation makes it seem likely that conjugate transpose was meant.) $\endgroup$
    – Michael E2
    Mar 20 at 0:55
  • $\begingroup$ Things are a bit confusing so that was my take on it. The author said in a comment that * was the complex conjugate and i assumed he meant plus transpose. Either way they will not be simultaneously diagonalizable. My answer was probably better set as a comment but it just didn’t fit. $\endgroup$
    – Mauricio de Oliveira
    Mar 20 at 1:48
  • $\begingroup$ @MauriciodeOliveira, I have updated my question. Indeed you verified that $[P, QR]=0$, so all I want are the simultaneous eigenvectors of $P$ and $QR$. $\endgroup$
    – Mike
    Mar 20 at 11:27

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