Let $A$ be a real invertible $n\times n$ matrix, and $B$ a real $n\times m$ matrix. $\mathrm{col}(.)$ denotes the column space of a matrix.
What are the conditions for $\mathrm{col}(AB) = \mathrm{col}(B)$ in terms of the eigenvectors of $A$?
2026-04-09 13:00:21.1775739621
$\mathrm{col}(AB) = \mathrm{col}(B)$
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Assume that $B$ has rank $m$. Then $\mathrm{rank}(AB)=m$ and $\mathrm{col}(AB)=\mathrm{col}(B)$ is equivalent to the existence of an $m\times m$ nonsingular matrix $M$ such that $AB=BM$. If $M$ is diagonalisable ($M=XDX^{-1}$ for some diagonal $D$ and nonsingular $X$), then $AB=BXDX^{-1}$ gives $A(BX)=(BX)D$ and hence the columns of $B$ are a basis of some invariant subspace of $A$ (formed from eigenvectors corresponding to nonzero eigenvalues of $A$).