If a matrix $A$ (it doesn't need to be a square matrix) has rank $r$, is it always possible to choose $r$ rows and $r$ columns in $A$ such that the matrix $B$ formed by removing the rows and columns that are not selected also has rank $r$, i.e., $B$ is invertible? If so, how can this be proven?
2026-04-04 00:51:14.1775263874
Does a matrix of rank $r$ always contain an invertible submatrix of size $r\times r$?
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If $A$ has rank $r$ it has at least $r$ rows. If it has more than $r$ rows, some of the rows can be formed by combining the other rows linearly, and can therefore be removed to form a matrix $A^-$ with only $r$ rows but also with the rank $r$. The same goes for the columns, so if $A^-$ has more than $r$ columns, some of them can be removed to form an $r$-by-$r$ matrix $B$, also of rank $r$.