Let $F$ be a field and let $A$ be a $n\times m$ matrix which has rank $r$. By the dot number 9 in the section Proposition, there exists an invertible $m\times m$ matrix $X$ and an invertible $n\times n$ matrix $Y$ such that $XAY=\left(\begin{smallmatrix}I_r&0\\0&0\end{smallmatrix}\right)$ in which $I_r$ denotes the $r\times r$ identity matrix. I wonder if there exists a matrix $P$ such that $PAP^{-1}=\left(\begin{smallmatrix}A_{11}&A_{12}\\0&0\end{smallmatrix}\right)$ in which $A_{11}$ has rank $r$.
2026-03-26 13:00:48.1774530048
The similarity of a $n\times m$ matrix
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First of all, you will need that $n=m$.
But even in such case, this is not true in general. Since $A$ is similar to a matrix with diagonal blocks $A_{11}$ and 0, then $A$ needs to have at exactly $n-r$ zero eigenvalues. The other eigenvalues (those of $A_{11}$) need to be be nonzero (otherwise $A_{11}$ will fail to be full rank).
So, this excludes certain matrices, namely those having more than $n-r$ zero eigenvalues.
For instance, if you consider the matrix $$A=\begin{bmatrix} 0 & 1\\0 & 0\end{bmatrix}.$$
The rank of this matrix is one but the term $A_{11}$ fails to be of rank one.
Your statement becomes generally true if you say that $[A_{11}\ A_{12}]$ has rank $r$.