Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 8F / Solution

For ordered bases $\mathcal{B}=(v_1,\dots ,v_n)$ of $V$ and $\mathcal{C}=(w_1,\dots ,w_m)$ of $W$, the matrix representation of a linear map $\gamma :V\to W$ is the matrix whose $j$th column consists of the $\mathcal{C}$-coordinates of $\gamma (v_j)$. Thus

The operators $\alpha ,\beta :V\to V$ are conjugate linear operators when there is an isomorphism $s:V\to V$ such thatIn one basis their matrices therefore satisfy $[\beta ]=[s{]}^{-1}[\alpha ][s]$, so they are similar matrices.

For invertible $\beta$, the mapis linear, and its inverse is $A\mapsto \beta A{\beta }^{-1}$; hence it is a linear isomorphism of $\mathrm{End}(V)$. If ${\beta }^{\prime }=s^{-1}\beta s$, define ${\Phi }_s(A)=s^{-1}As$. A direct substitution givesso ${\varphi }_{{\beta }^{\prime }}$ and ${\varphi }_{\beta }$ are conjugate. This is the conjugation operator on an endomorphism space.

It remains to compute the Jordan normal form over $\mathbb{C}$. By the preceding conjugacy, we may put $\beta$ in Jordan form.

Ifthen the matrix units $E_{ij}$ are eigenvectors becauseThusThis includes the scalar case $\lambda =\mu$, when ${\varphi }_{\beta }$ is the identity.

Otherwise $\beta$ has one size-two Jordan block. Multiplying $\beta$ by a nonzero scalar does not change ${\varphi }_{\beta }$, and conjugating within its Jordan class allows us to use $J=I+N$ with $N=E_{12}$. Put $D={\varphi }_J-I$. Direct multiplication givesHence $D^3=0$, $\mathrm{rank}D=2$, and $\mathrm{rank}{D}^2=1$. The nilpotent Jordan blocks of $D$ therefore have sizes three and one. Adding the identity givesThese two cases are the Jordan normal form of conjugation on two-by-two matrices.

Solved by gpt-5.6-sol high.