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IMC / 2003 / Problems / Day 1, P3

IMC 2003 · Day 1 · P3

medium
linear algebraworth 20 pts

Let AA be an n×nn \times n real matrix such that 3A3=A2+A+I3 A^3 = A^2 + A + I (II is the identity matrix). Show that the sequence AkA^k converges to an idempotent matrix. (A matrix BB is called idempotent if B2=BB^2 = B.)

Solution (official)

The minimal polynomial of AA is a divisor of 3x3x2x13 x^3 - x^2 - x - 1. This polynomial has three different roots. This implies that AA is diagonalizable: A=C1DCA = C^{-1} D C where DD is a diagonal matrix. The eigenvalues of the matrices AA and DD are all roots of polynomial 3x3x2x13 x^3 - x^2 - x - 1. One of the three roots is 1, the remaining two roots have smaller absolute value than 1. Hence, the diagonal elements of DkD^k, which are the kkth powers of the eigenvalues, tend to either 0 or 1 and the limit M=limDkM = \lim D^k is idempotent. Then limAk=C1MC\lim A^k = C^{-1} M C is idempotent as well.

How the field did

contestants scored
185
average (of 20)
11.64
solved (≥ 80%)
44.9%
near-0 (≤ 10%)
24.9%
discrimination
0.57

Score distribution (field cohort)

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

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