Let $r=\dim\operatorname{im}T$. Since the <kernel of a linear map> and its <image> agree, the <rank-nullity theorem> gives $\dim V=r+r$. Thus \b[$\dim V=2r$ is even]. Every $Tv$ lies in $\ker T$, so $T^2=0$. The <minimal polynomial> therefore divides $X^2$. It cannot be $1$, and it cannot be $X$ since $T\ne0$; hence \b[the minimal <polynomial> is $X^2$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1 / 1G / 1 / Solution

 Solution (source code)