IMC / 2000 / Problems / Day 1, P3

IMC 2000 · Day 1 · P3

medium

$A$ and $B$ are square complex matrices of the same size and $\operatorname{rank}(AB - BA) = 1.$ Show that $(AB - BA)^2 = 0$ .

Solution (official)

Let $C = AB - BA$ . Since $\operatorname{rank} C = 1$ , at most one eigenvalue of $C$ is different from 0. Also $\operatorname{tr} C = 0$ , so all the eigevalues are zero. In the Jordan canonical form there can only be one $2 \times 2$ cage and thus $C^2 = 0$ .

How the field did

contestants scored

114

average (of 20)

10.71

solved (≥ 80%)

44.7%

near-0 (≤ 10%)

33.3%

discrimination

0.58

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

IMC 2000 · Day 1 · P3

How the field did

Score distribution (field cohort)

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