IMC 2007 · Day 1 · P3

The possible values for $k$ are 1 and 2.

If $k = 1$ then $P(x) = \alpha x^2$ and we can choose $A_1 = \left( \begin{smallmatrix} 1 & 0 \\ 0 & \alpha \end{smallmatrix} \right)$.

If $k = 2$ then $P(x, y) = \alpha x^2 + \beta y^2 + \gamma xy$ and we can choose matrices $A_1 = \left( \begin{smallmatrix} 1 & 0 \\ 0 & \alpha \end{smallmatrix} \right)$ and $A_2 = \left( \begin{smallmatrix} 0 & \beta \\ -1 & \gamma \end{smallmatrix} \right)$.

Now let $k \ge 3$. We show that the polynomial $P(x_1, \dots, x_k) = \sum\limits_{i=0}^{k} x_i^2$

is not good. Suppose that $P(x_1, \dots, x_k) = \det \left( \sum\limits_{i=0}^{k} x_i A_i \right)$. Since the first columns of $A_1, \dots, A_k$ are linearly dependent, the first column of some non-trivial linear combination $y_1 A_1 + \dots + y_k A_k$ is zero. Then $\det(y_1 A_1 + \dots + y_k A_k) = 0$ but $P(y_1, \dots, y_k) \ne 0$, a contradiction.