Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2 / 8E / b

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.