3
$\begingroup$

I'm looking for transformations (isometries or not) that leave the spectrum of a matrix unchanged. I've only been able to come up with

  • similarity transformations and
  • transposition (is transposition in fact just a similarity transformation?).

What are some others? Thanks.

  • 0
    Transposition leaves not just the spectrum but the Jordan normal form unchanged, so it is in fact a similarity transform. (I suspect most things worthy of being called "transformations" would have the same property if they preserve the raw spectrum). – 2017-02-24
  • 0
    It is almost surely the only transformations that leave unchanged the spectrum. – 2017-02-24
  • 0
    @HenningMakholm So you're saying that we can write transposition of a matrix $A\in\mathbb{R}^{n\times n}$ like $A^T=P^{-1}AP$ for some invertible $P$, or is it only if $A\in\mathbb{C}^{n\times n}$? – 2017-02-24
  • 0
    @JeanMarie This is also my feeling, but can it be seen in a more rigorous way? – 2017-02-24
  • 0
    @Lovsovs: The Jordan-form argument gives you a $P\in \mathbb C^{n\times n}$ that does the trick. Now, real matrices that are similar over $\mathbb C$ [are also similar over $\mathbb R$](https://math.stackexchange.com/questions/57242/similar-matrices-and-field-extensions). – 2017-02-24
  • 0
    Let $A$ and $B$ two square matrices with same eigenvalues. Then it can be shown that there is a nonsingular $P$ such that $$A=P^{-1}BP$$ the same way the svd theorem is proved – 2017-02-24
  • 0
    @polfosol Huh, I didn't know the implication went the other way as well. This answers my question. If you post this as an answer with either a heuristic argument for its truth or a reference containing the proof, I'll gladly accept. – 2017-02-24
  • 0
    @HenningMakholm Thank you! A last question: Is it correct that sim.transforms are, in general, not isometries, but can be? – 2017-02-24
  • 1
    @polfosol: How about $A=({}^1_0\;{}^1_1)$ and $B=({}^1_0\;{}^0_1)$? – 2017-02-24
  • 0
    @HenningMakholm Well, that was unexpected `:p` – 2017-02-24
  • 0
    Um, wait a moment -- I suppose a "similarity transformation" ought to mean $A\mapsto P^{-1}AP$ for some _fixed_ $P$. My argument doesn't show that transposition is of _that_ form, only that _for each $A$_ there is such a $P$. – 2017-02-24
  • 0
    @HenningMakholm Yes, thank you, that was clear (in my head, not the way I wrote it). – 2017-02-24
  • 2
    Indeed there cannot be any $P$ such that $A^T=P^{-1}AP$ for all $A$. The only $P$s that work for $A=({}^1_0\;{}^0_2)$ will be diagonal matrices, but there is no diagonal $P$ that works for $A=({}^1_0\;{}^1_1)$ because we need to get $A^T=({}^1_1\;{}^0_1)$ but a product of upper triangular matrices is upper triangular. – 2017-02-24
  • 0
    @HenningMakholm: Also $A\mapsto PAP^{-1}$ is an algebra homomorphism, $A\mapsto A^T$ is an "anti-homomorphism," because $(AB)^T=B^TA^T$. – 2017-02-24

1 Answers 1

2

If the transformation is linear and the underlying field is algebraically closed, it is known that all possible choices are in the form of $A\mapsto P^{-1}AP$ or $A\mapsto P^{-1}A^\top P$, where $P$ is any constant but invertible matrix.

For other fields, you should first define what does it mean by "leaving the spectrum unchanged", since a matrix may not have a full spectrum over the ground field. Anyway, you may search for any known results on the internet or on any journal database. Some relevant keywords are "spectrum preserving linear map" and "linear preservers of spectrum/eigenvalues".