Let $A,B \in M_{n}(\mathbb{R}).$ Show that $rank(A) = rank(B)$ if and only if there exist nonsingular matrices $X,Y,Z \in M_{n}(\mathbb{R})$ such that \[ AX + YB = AZB. \]
Problem
Source: Romania National Olympiad 2023
Tags: matrix, linear algebra, Non singular matrices
14.04.2023 09:14
23.05.2023 17:53
Here $K$ is a field with more than $2n$ elements. Assume that $rank(A)=rank(B)$; then there are $P,Q\in GL_n(K)$ s.t. $A=PBQ$. Thus, for every $a\in K$, $A(1/2B+aQ^{-1})+(1/2A-aP)B=AB$. $\det(1/2B+aQ^{-1})$ and $\det(1/2A-aP)$ are polynomials of degree $n$ in $a$. Then there is $a\in K$ s.t. $X=1/2B+aQ^{-1}\in GL_n$ and $Y=1/2A-aP\in GL_n$. Choosing $Z=I_n$, we are done. $\square$ EDIT. In fact $rank(A)=rank(B)$ iff There exist non-singular matrices $X,Y\in GL_{n}(\mathbb{K})$ such that $AX + YB = AB$.
23.05.2023 21:50
loup blanc wrote: Here $K$ is a field with more than $2n$ elements. Assume that $rank(A)=rank(B)$; then there are $P,Q\in GL_n(K)$ s.t. $A=PBQ$. Thus, for every $a\in K$, $A(1/2B+aQ^{-1})+(1/2A-aP)B=AB$. $\det(1/2B+aQ^{-1})$ and $\det(1/2A-aP)$ are polynomials of degree $n$ in $a$. Then there is $a\in K$ s.t. $X=1/2B+aQ^{-1}\in GL_n$ and $Y=1/2A-aP\in GL_n$. Choosing $Z=I_n$, we are done. $\square$ EDIT. In fact $rank(A)=rank(B)$ iff There exist non-singular matrices $X,Y\in M_{n}(\mathbb{K})$ such that $AX + YB = AB$. If you write $A(aQ^{-1})+(A-aP)B=AB$ you can see that the result is true when $K$ has more than $n+1$ elements (and you can avoid the problem with characteristic 2)