MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We all know that a set of commuting diagonalizable matrices can be simultaneously put in diagonal form. My general question is:

Under what conditions can a set of (diagonalizable) matrices be simultaneously put in monomial form?

That is, when can find a basis with respect to which a collection of operators $\{A_i\}$ take the form $D_iP_i$ where $D_i$ is a diagonal matrix and $P_i$ is a permutation matrix? Clearly they must be diagonalizable, but I would like sufficient conditions.

For example:

Is it enough if (a) there exist $n_i$ so that $A_i^{n_i}$ pairwise commute and are diagonalizable? Or (b) if the set $\{A_i\}$ is closed under conjugation, i.e. $A_iA_jA_i^{-1}=A_{i(j)}$ (which obviously implies (a))?

Edit: As Will Sawin points out, this is not enough as stated, so let me add another condition to (b): suppose that $|\{A_i\}|\leq n$ where $A_i\in \mathbb{C}^{n\times n}$, so the number of generators is less than the dimension of the representation.

As you might guess this has something to do with representations of knot groups, but I am interesting in the general problem.

share|cite|improve this question

$(a)$ and $(b)$ are both true for finite group representations, but some of them, such as the $4$-dimensional representation of $A_5$, cannot be put in this form.

For $(a)$, the group generated by the matrices need not even be virtually abelian. Take $A_1 = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$ and $A_2 = \left(\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right)$. Take $n_1=4$ and $n_2=3$. Then $A_1^{n_1}=A_2^{n_2}=I$, but the group generated by $A_1$ and $A_2$ is $SL_2(\mathbb Z)$, not virtually abelian. Not sure about $b$.

I can't think of a better condition.

share|cite|improve this answer
Thanks Will! Including (a) was an afterthought that should have been removed upon further reflection. – Eric Rowell Nov 13 '12 at 16:22
One potential problem with the new condition is that every finite group representation satisfies it when you add enough trivial summands. I can't immediately think of an example of a finite group representation that cannot be put in this form after adding enough trivial summands, but it seems likely that one exists. – Will Sawin Nov 13 '12 at 17:04
I think the automorphism group of a nice lattice, like the $E_8$ lattice or the Leech lattice, with the standard representation, should give a counterexample. – Will Sawin Nov 13 '12 at 17:42

As Will Sawin suggests, the answer to the last version of the question is negative. If we tak $G$ to be a non-Abelian finite simple group which is not a doubly transitive permutation group, and $\chi$ to be a non-trivial complex irreducible character of last degree o $G,$ then it is impossible to write $\chi$ + any multiple of the trivial character as a sum of characters each induced from linear characters of (not necessarily proper) subgroups. For if $\lambda$ is a non-trivial linear character of a proper subgroup $H$ of $G,$ then ${\rm Ind}_{H}^{G}(\lambda)$ does not contain the trivial character by Frobenius reciprocity. Neither can it be $\chi$, for otherwise ${\rm Ind}_{H}^{G}(1)$ would have a trivial constituent, and at least one non-trivial irreducible constituent $\mu$. But then $\mu(1) < \chi(1)$, contrary to the choice of $\chi.$ On the other hand,if $H$ is a proper subgroup of $G$, the permutation character ${\rm Ind}_{H}^{G}(1)$ only contains the trivial character once, but it can't be $1 + \chi,$ for otherwise $G$ would be a doubly transitive permutation character on the cosets of $H$ in $G,$ contrary to the choice of $G.$

share|cite|improve this answer

There seems to be a stronger condition than your (b) (the original one, with a closed set), i.e.:

Two invertible diagonalizable matrices $X, Y$ are simultaneously monomializable (in your sense) iff $Y X Y^{-1}$ and $X$ commute (i.e. are diagonalizable in the same basis) and have the same spectrum (including degeneracies).


Let $X = C A C^{-1}$ and $Y=CPBC^{-1}$, where $A$ and $B$ are diagonal matrices and $P$ is a permutation matrix.

Then $$Y X Y^{-1} = C PBAB^{-1}P^{-1}C^{-1} = CPAP^{-1}C^{-1},$$ so both parts are diagonal in the basis $C$, with permuted diagonal elements.


Let $X=CAC^{-1}$ and $Y X Y^{-1} = C P A P^{-1} C^{-1}$, again, for $A$ diagonal and $P$ - a permutation matrix.

Then $$C^{-1}YCAC^{-1}Y^{-1}C=PAP^{-1}.$$ Now, let's define $B = P^{-1}C^{-1}YC$, which converts the above in the commutation condition $BAB^{-1}=A$. Hence*), $B$ is diagonal (as $A$ is diagonal).

So, we have $PB = C^{-1}YC$, what we wanted to show.

*) There is one fragile part of which I'm aware of: if some eigenvalues are degenerate, then there is freedom in choice of $C$. So then $B$ needs not to be diagonal, just there is a matrix $D$ (non diagonal), commuting with $A$, such that $D^{-1} BD$ is diagonal.

(I'm posting it anyway, as perhaps there is a simple fix of that.)

share|cite|improve this answer
In "$\Rightarrow$" you assume there is a basis with respect to which $X$ is diagonal and $Y$ is a monomial matrix. This is stronger than "simultaneously monomializable." Indeed, take $X,Y$ to be the $3\times 3$ permutation matrices corresponding to $(1\/2)$ and $(1\/2\/3)$. Then they are monomial in the standard basis, but $YXY^-1$ corresponds to $(2\/3)$, which does not commute with $(1\/2)$. – Eric Rowell Nov 20 '12 at 17:28
@Eric Thanks. (I see, I had in mind a different, stronger condition.) – Piotr Migdal Nov 21 '12 at 16:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.