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IMC / 1994 / Problems / Day 1, P4

IMC 1994 · Day 1 · P4

linear algebraworth 18 pts

Let αR{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 FGGF=αFF \circ G - G \circ F = \alpha F.

a) Show that for all kNk \in \mathbb{N} one has FkGGFk=αkFkF^k \circ G - G \circ F^k = \alpha k F^k.

b) Show that there exists k1k \ge 1 such that Fk=0F^k = 0.

Solution (official)

For a) using the assumptions we have FkGGFk=i=1k(Fki+1GFi1FkiGFi)==i=1kFki(FGGF)Fi1==i=1kFkiαFFi1=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)=FGGFL(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 Fk0F^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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