Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 1

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

Write $N=T-I$. Cayley-Hamilton theorem gives $N^3=0$. Its minimal polynomial is therefore $X^k$ for exactly one $k\in \{1,2,3\}$, so the possibilities for that of $T$ areLet $d=\dim \ker N$, which is the dimension of the only eigenspace. If $d=3$, then $N=0$ and $T=I$.

If $d=2$, rank-nullity theorem gives $\dim \mathrm{im}N=1$. The restriction of the nilpotent map $N$ to this invariant line is zero, so $N^2=0$. Choose $v_3$ with $v_2=N{v}_3\ne 0$, and choose $v_1\in \ker N$ independent of $v_2$. Then $v_3\notin \ker N$, so these three vectors form a basis; its matrix for $T$ is the printed $J_1(1)\oplus J_2(1)$.

If $d=1$, then $\mathrm{im}N$ has dimension two, so $N^2$ cannot vanish: otherwise $\mathrm{im}N\subseteq \ker N$. Choose $v_3$ with $N^2{v}_3\ne 0$ and set $v_2=N{v}_3$, $v_1=N^2{v}_3$. Applying $N^2$ and then $N$ to a putative linear relation proves these three vectors independent. The matrix is the printed $J_3(1)$.

These cases exhaust the possible eigenspace dimensions. They are pairwise dissimilar because similarity preserves the eigenspace dimension, equivalently the three different minimal polynomials. Hence every matrix is similar to exactly one of the three displayed forms.