Past exam of the linear algebra course of the University of Oxford / 2017 / 1 / a

For a linear transformation $T$ of a finite-dimensional vector space over a field $F$, its minimal polynomial $m_T$ is the unique monic polynomial of least degree satisfying $m_T(T)=0$. There is at least one nonzero polynomial with this property because the powers $I,T,T^2,\dots$ lie in the finite-dimensional vector space of linear maps $V\to V$. Dividing any other such polynomial by $m_T$ leaves a remainder of smaller degree that also vanishes at $T$, so that remainder must be zero. This also proves uniqueness.

If $C$ represents $T$ in any basis, the paper's characteristic polynomial isIt is independent of the basis, since changing the basis replaces $C-xI$ by a similar matrix, preserving its determinant. Its degree is $\dim V$ and its leading coefficient is $(-1{)}^{\dim V}$; $m_T$ is always monic.

The Cayley-Hamilton theorem statesLet $d=\dim V$. Extend the field to an algebraically closed field and choose a basis in which the matrix of $T$ is an upper triangular matrix, as permitted. Write its diagonal entries as ${\lambda }_1,\dots ,{\lambda }_d$ and put $V_i=\mathrm{span}(e_1,\dots ,e_i)$. Each $V_i$ is an invariant subspace, andApply $T-{\lambda }_{d}I$ first, then $T-{\lambda }_{d-1}I$, and continue. The resulting product sends $V_d$ into $V_0=\{0\}$. These factors are commuting operators, soThe determinant of an upper triangular matrix gives ${\chi }_T(x)=(-1{)}^d{\prod }_i(x-{\lambda }_i)$. Thus ${\chi }_T(T)=0$ after the field extension. Its entries were in the original field, whose embedding is injective, so they were already zero there. The zero-dimensional case is immediate as well.