Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

The title explains it all. Can every square matrix $A$ be written as $A=B_1B_2=B_2B_1$ of any two matrices $B_1$,$B_2$.

share|cite|improve this question
$B_1=A$ and $B_2=I$? – Paul Feb 1 '13 at 12:11
Or $B_1 = \lambda^{-1} A$ and $B_2 = \lambda I$ for some scalar $\lambda \neq 0$. – Mikko Korhonen Feb 1 '13 at 12:12
good one! but that is kind of trivial. If $A$ is diagonalizable, I could write $B_1=X\Lambda^{p} X^{-1}$ and $B_2=X\Lambda^{q} X^{-1}$ where $A=X\Lambda X$ and $p+q=1$. That is a bit non-trivial – dineshdileep Feb 1 '13 at 12:13
Well, you did say "any two matrices". Do you want the $B_1$ and $B_2$ to have some particular property? – Mikko Korhonen Feb 1 '13 at 12:15
If $A$ is invertible, you can write it as $A^{73}A^{-72}=A^{-72}A^{73}$. – Gerry Myerson Feb 1 '13 at 12:29
up vote 1 down vote accepted

The answer is yes, and you can produce arbitrarily many such factorisations. Take any polynomial $P$ whose zeros are not eigenvalues of $A$, then $P(A)$ is invertible, so you can write $A=AP(A)\cdot P(A)^{-1}$. More generally, you can replace $P$ by an entire function (or power series that converges on a disk with radius bigger than the norm of $A$).

I think you have to add more details to your question, e.g. require $B_1$ and $B_2$ to satisfy some additional conditions.

Here's another question: Can you characterise all possible factorisations?

share|cite|improve this answer
Since every square matrix satisfies a polynomial equation (Cayley-Hamilton gives you a concrete one), what is the point of using entire functions or power series? You get nothing more than with polynomials of degree up to $n$ (the size of the matrix). – Marc van Leeuwen Feb 1 '13 at 15:04

To answer a variant of the question in a comment by OP under the question: a diagonal matrix with distinct entries cannot be written as a product of two commuting non-diagonal matrices (in fact both commuting matrices need to be diagonal). If $A=BC=CB$ then $B$ commutes with $BC=A$ (similarly for $C$), and any matrix that commutes with $A$ must stabilise each of the eigenspaces of $A$; since here these are $1$-dimensional, this means $B$ and $C$ are diagonal matrices.

share|cite|improve this answer

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.