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Let $A,B \in \mathbb R^{n \times n}$.

$A$ and $B$ are similar if there exists $P \in GL(n, \mathbb R)$ such that $AP=PB$.

While we could define something like we can define something like

$A$ and $B$ are 'conjugate' if there exists $P \in \mathbb R^{n \times n}$ such that $AP=PB$,

this would be kind of pointless since we could always pick $P=0$. Of course there are definitions for $A$ and $B$ to be 'conjugate in $X$' for some $X \subseteq \mathbb R^{n \times n}$ (or $X \subseteq GL(n, \mathbb R)$) if there exists $P \in X$ such that $AP=PB$, like here.

Question 1: Is it possible that $AP=PB$ for some nonzero yet non-invertible $P$? I have a feeling I'm missing some obvious counterexample. If no, then please help me prove that $P$ must either be zero or invertible.

Question 2: If yes, then I have a feeling there are infinitely many such $P$'s for any given $A$ and $B$. Is it true that for every $A$ and $B$, there exists such a $P$?

  • Context: Bullet 3.1 here
José Carlos Santos
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BCLC
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  • My impression is that in linear algebra literature, people seldom write "$A$ is conjugate to $B$". When they do so, they almost always mean conjugation in a group-theoretic sense. Hence "conjugate" simply means "similar". Your definition of conjugation without requiring $P$ to be invertible seems unusual. – user1551 Feb 13 '20 at 09:43
  • @user1551 I'm not actually defining it. I'm just saying that if we did do something like that, then 'this would be kind of pointless'...Or wait are you referring to the question I linked to ? – BCLC Feb 13 '20 at 09:44

2 Answers2

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As the other answer shows, it is possible that $AP=PB$ for some nonzero yet non-invertible $P$, and it is also possible to find matrices $A,B$ that a non-invertible, non-zero solution $P$ exists, yet no invertible solution $P$ exists.

If yes, then I have a feeling there are infinitely many such $P$'s for any given $A$ and $B$. Is it true that for every $A$ and $B$, there exists such a $P$?

The answer is no; but we can be a bit more thorough than that. The $P$'s that you are looking for are the solutions to the Sylvester equation $$ AP + P(-B) = 0. $$ This equation will have infinitely many solutions if and only if $A$ and $B$ have a common eigenvalue. More specifically, if $\lambda \in \Bbb C$ is such that $Ax = \lambda x$ and $B^Ty = \lambda y$, then the matrix $P = xy^T$ (and its multiples) are solutions to this equation.

Of course, an invertible solution $P$ exists if and only if $A,B$ are similar. So, a necessary (but insufficient) condition for the existence of an invertible solution is that $A$ and $B$ have identical eigenvalues with identical (algebraic) multiplicites. If all eigenvalues have multiplicity $1$, then the condition also becomes sufficient.

Ben Grossmann
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    @John A generalization of your "non-invertible conjugacy" comes up in abstract algebra and category theory in the form of natural transformation as opposed to natural isomorphism. – Ben Grossmann Feb 13 '20 at 09:47
  • Thanks Omnomnomnom! I have referenced your answer in this question. Please consider answering or commenting on said question in re the (homogeneous) Sylvester equation, if you think it will be helpful. – BCLC Feb 13 '20 at 09:53
  • Wait it says 'A Sylvester equation has a unique solution for X exactly when there are no common eigenvalues of A and −B.' when $A$ and $-B$ have a common eigenvalue, there should be either no $P$ and infinitely many $P$'s. Why is there necessarily some $P$ when $A$ and $-B$ have a common eigenvalue please? – BCLC Feb 13 '20 at 09:58
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    @JohnSmithKyon I address this in my answer. Adapting the sentence to the wiki version: more specifically, if $\lambda \in \Bbb C$ is such that $Ax = \lambda x$ and $B^Ty = -\lambda y$, then the matrix $P = xy^T$ (and its multiples) are solutions to this equation. – Ben Grossmann Feb 13 '20 at 10:32
  • Oh right yeah. Thanks. – BCLC Feb 13 '20 at 10:44
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If $A=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$, $B=\left[\begin{smallmatrix}2&0\\0&0\end{smallmatrix}\right]$, and $P=\left[\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right]$, then $AP=PB$. However, $P\neq\left[\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right]$ and $A$ and $B$ are not similar.

José Carlos Santos
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