IMC / 2011 / Problems / Day 1, P2

IMC 2011 · Day 1 · P2

Does there exist a real $3 \times 3$ matrix $A$ such that $\operatorname{tr}(A) = 0$ and $A^2 + A^t = I$ ? ( $\operatorname{tr}(A)$ denotes the trace of $A$ , $A^t$ is the transpose of $A$ , and $I$ is the identity matrix.)

(Moubinool Omarjee, Paris)

Solution (official)

The answer is NO.

Suppose that $\operatorname{tr}(A) = 0$ and $A^2 + A^t = I$ . Taking the transpose, we have $A = I - (A^2)^t = I - (A^t)^2 = I - (I - A^2)^2 = 2A^2 - A^4,$ $A^4 - 2A^2 + A = 0.$ The roots of the polynomial $x^4 - 2x^2 + x = x(x-1)(x^2 + x - 1)$ are $0, 1, \frac{-1 \pm \sqrt{5}}{2}$ so these numbers can be the eigenvalues of $A$ ; the eigenvalues of $A^2$ can be $0, 1, \frac{1 \pm \sqrt{5}}{2}$ .

By $\operatorname{tr}(A) = 0$ , the sum of the eigenvalues is 0, and by $\operatorname{tr}(A^2) = \operatorname{tr}(I - A^t) = 3$ the sum of squares of the eigenvalues is 3. It is easy to check that this two

conditions cannot be satisfied simultaneously.

IMC 2011 · Day 1 · P2

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