On linear operators

Let \alpha \in \mathbb{R} \setminus \{0\} and suppose that F ,G are linear operators from \mathbb{R}^n into \mathbb{R}^n satisfying

(1)   \begin{equation*}F\circ G - G \circ F =\alpha F \end{equation*}

  1. Show that for all k \in \mathbb{N} one has

        \[F^k \circ G - G \circ F ^k= \alpha k F^k\]

  2. Show that there exists k \geq 1 such that F^k =0.

Solution

  1. Using the assumptions we have

    \begin{aligned} F^k \circ G - G \circ F^k &= \sum_{i=1}^{k} \bigg (F^{k-i+1} \circ G \circ F^{i-1} - F^{k-i} \circ G \circ F^i \bigg ) \\ &= \sum_{i=1}^{k} \bigg( F^{k-i} \circ \left ( F \circ G- G \circ F \right ) \circ F^{i-1} \bigg)\\ &= \sum_{i=1}^{k} F^{k-i} \circ \alpha F \circ F^{i-1} \\ &= \alpha k F^k \end{aligned}

  2. Consider the linear operator \mathcal{L}(F) = F\circ G - G \circ F acting over all n \times n matrices F. It may have at most n^2 different eigenvalues. Assuming that F^k \neq 0 for every k we get that \mathcal{L} has infinitely many different eigenvalues \alpha k in view of (i). This is a contradiction.

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