We know that any $n \times n$ matrix $X$ with $rank(X)=1$ can be written $u(^Tv)$ where $u, v$ are vectors and $^TA$ is the transpose of $A$. This leads to $X^2=u(^Tv)u(^Tv)=u(^Tvu)(^Tv)=cu(^Tv)$, where $c$ is the scalar product of $u$ and $v$. We can see that $tr(X)=tr(u(^Tv))=tr((^Tv)u)=c$. Thus $X^2=tr(X)X$. This looks very similar to the characteristic polynomial of a $2\times2$ matrix, with the determinant $0$. Is $x^2-tr(X)x$ the minimal polynomial for an arbirary matrix with $rank(X)=1$? And can we generalize this for $rank(X)=k < n$? It seems intuitive that if $rank(X)=2$ then $X^3-tr(X)X^2+\frac{tr(X)^2-tr(X^2)}{2}X$ is $0$. Can this be generalized?
2026-04-06 17:50:16.1775497816
Is it possible to find the minimal polynomial of a matrix knowing its rank?
137 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in LINEAR-ALGEBRA
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