Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1 / 1G (Linear Algebra)

Let $r=\dim \mathrm{im}T$. Since the kernel of a linear map and its image agree, the rank-nullity theorem gives $\dim V=r+r$. Thus $\dim V=2r$ 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 $T\ne 0$; hence the minimal polynomial is $X^2$.

Choose $0\ne v\in A_3$ and write $v=a_1+a_2$ using the direct sum of vector spaces $V=A_1\oplus A_2$. Neither component is zero: otherwise $v$ would lie in one of the zero intersections $A_3\cap A_1$ or $A_3\cap A_2$. They are linearly independent, so $W=\mathrm{span}\{a_1,a_2\}$ is a two-dimensional vector subspace. Uniqueness of the direct-sum decomposition gives $W\cap A_1=\mathrm{span}\{a_1\}$ and $W\cap A_2=\mathrm{span}\{a_2\}$. Also $v\in W\cap A_3$, but $W$ cannot be contained in $A_3$ because $a_1\in A_1\setminus \{0\}$. Hence $W\cap A_3$ is one-dimensional. This constructs the required common plane for three pairwise complementary subspaces.