Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2 / 8E (Linear Algebra)

A direct determinant calculation givesMoreover,has rank one, so its kernel has dimension two, while $(A-I{)}^2=0$ and $A\ne I$. Hence the minimal polynomial isThere are two Jordan blocks, because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus

Choose the least $k\ge 1$ such thatThenis nonzero by minimality, andTherefore $fw=\alpha w$, so $w$ is the required nonzero eigenvector.

We use induction on $d$. SupposeApplying $f-{\alpha }_{d}I$ givesBy induction, every $c_i({\alpha }_i-{\alpha }_d)=0$ for $i<d$. The eigenvalues are distinct, so $c_i=0$ for $i<d$, and the original relation then gives $c_d=0$. Thus eigenvectors with distinct eigenvalues are linearly independent.

If $\alpha \ne \beta$, the operator $f-\beta I$ is invertible on $W_{\alpha ,n}$. Indeed, there it equalswhose inverse is the finite geometric series

Now use induction on $d$. If ${\sum }_{i=1}^d{v}_i=0$, apply $(f-{\alpha }_{d}I{)}^n$. The $v_d$ term vanishes, while every transformed vectoris nonzero by the invertibility just proved and still lies in $W_{{\alpha }_i,n}$. The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the generalized eigenspaces for distinct eigenvalues form a direct sum, so $v_1,\dots ,v_d$ are linearly independent.