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

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$B_1=A$ 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
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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

2 Answers 2

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?

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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.

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