Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 8G (Linear Algebra)

The characteristic polynomial isTo prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb{C}$, ${\chi }_A$ has a root $\lambda$, so $A$ has an eigenvector $v_1$. Extend it to a basis. In this basis,By induction, a change among the remaining $n-1$ basis vectors makes $B$ upper triangular. Thus the triangularization over an algebraically closed field gives

The minimal polynomial $m_A$ is the monic polynomial of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular matrix similar to $A$, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, and putThenThe factors $T-{\lambda }_{k}I$ commute, so applying all of them successively lowers the invariant flag to zero:Similarity gives the same polynomial identity for $A$. Hence a nonzero monic annihilating polynomial of degree $n$ exists, and a least-degree one exists.

If $m$ and $\tilde{m}$ were two monic annihilating polynomials of the same least degree, then $m-\tilde{m}$ would be an annihilating polynomial of smaller degree unless it were zero. Thus the minimal polynomial is unique, and the minimal polynomial bound from a triangular invariant flag gives

The eigenvalue equation $D_{r}f=\lambda f$ isIf $\lambda =-1$, this forces $f(x+r)=0$ for every $x$, hence $f=0$, so $-1$ is not an eigenvalue. If $c=1+\lambda \ne 0$, choose arbitrary values on representatives of the cosets of $r\mathbb{Z}$ and extend byThis produces nonzero eigenfunctions. Choosing functions supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Thereforeand every eigenspace is infinite-dimensional.

Direct expansion giveswhich is symmetric in $r,s$. Hence

Suppose a degree-$n$ polynomial $p$ were a sum of $n$ periodic functions,Apply the commuting product $D_{r_1}\cdots D_{r_n}$. Every term on the right is killed by its corresponding factor, whereas for leading coefficient $a_n$ the mixed finite difference of a polynomial givesThis contradiction proves