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IMC / 1994 / Problems / Day 1, P4

IMC 1994 · Day 1 · P4

linear algebraworth 18 pts

Let α∈R∖{0}\alpha \in \mathbb{R} \setminus \{0\} and suppose that FF and GG are linear maps (operators) from Rn\mathbb{R}^n into Rn\mathbb{R}^n satisfying F∘G−G∘F=αFF \circ G - G \circ F = \alpha F.

a) Show that for all k∈Nk \in \mathbb{N} one has Fk∘G−G∘Fk=αkFkF^k \circ G - G \circ F^k = \alpha k F^k.

b) Show that there exists k≥1k \ge 1 such that Fk=0F^k = 0.

Solution (official)

For a) using the assumptions we have Fk∘G−G∘Fk=∑i=1k(Fk−i+1∘G∘Fi−1−Fk−i∘G∘Fi)==∑i=1kFk−i∘(F∘G−G∘F)∘Fi−1==∑i=1kFk−i∘αF∘Fi−1=αkFk.\begin{align*} F^k \circ G - G \circ F^k &= \sum_{i=1}^{k} \left( F^{k-i+1} \circ G \circ F^{i-1} - F^{k-i} \circ G \circ F^{i} \right) = \\ &= \sum_{i=1}^{k} F^{k-i} \circ (F \circ G - G \circ F) \circ F^{i-1} = \\ &= \sum_{i=1}^{k} F^{k-i} \circ \alpha F \circ F^{i-1} = \alpha k F^k. \end{align*} b) Consider the linear operator L(F)=F∘G−G∘FL(F) = F \circ G - G \circ F acting over all n×nn \times n matrices FF. It may have at most n2n^2 different eigenvalues. Assuming that Fk≠0F^k \ne 0 for every kk we get that LL has infinitely many different eigenvalues αk\alpha k in view of a) – a contradiction.

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