IMC 2022 · Day 2 · P7

Hint: Consider the trace and the rank of $A$.

Lemma. For any idempotent matrix $B$ $$\operatorname{tr}(B) = \operatorname{rank}(B)$$ Proof. Observe that an idempotent matrix satisfies the equation $\lambda(1 - \lambda) = 0$. Hence the minimal polynomial is a product of linear factors and the matrix is diagonalizable. Therefore, the rank of the matrix equals the number of non-zero eigenvalues. Since the matrix has eigenvalues 0 or 1, this provides that the trace is equal to the number of unity eigenvalues, or non-zero eigenvalues.

It can be shown that $\sum\limits_{i=1}^{k} A_i$ is also an idempotent. Indeed, $$\left( \sum_{i=1}^{k} A_i \right)^2 = \sum_{i=1}^{k} A_i^2 + \sum_{i \ne j} (A_i A_j + A_j A_i) = \sum_{i=1}^{k} A_i$$ Applying the lemma one can obtain $$\sum_{i=1}^{k} \operatorname{rank}(A_i) = \sum_{i=1}^{k} \operatorname{tr}(A_i) = \operatorname{tr} \left( \sum_{i=1}^{k} A_i \right) = \operatorname{rank} \left( \sum_{i=1}^{k} A_i \right) \leqslant n$$ The required inequality follows.

We first prove that for idempotents $A, B$ with $AB = -BA$ we already must have $AB = BA = 0$. Indeed, it is clear that $ABx = BAx = 0$ for $x \in \ker(A)$ so it suffices to prove the same for $x \in \operatorname{im}(A)$, i.e. when $Ax = x$. But then writing $Bx = y$ we have $Ay = -y$ i.e. $y = -Ay = -A^2 y = Ay = -y$ and hence $y = 0$ so that again $ABx = BAx = 0$.

Henceforth, we can assume the stronger condition $A_i A_j = 0$ for all $i \ne j$. We next claim that all the image spaces $V_i$ of $A_i$ are linearly independent. This will imply the claim, since then the sum of their dimensions can be at most $n$, and so one of them has to be $\le \frac{n}{k}$. Now, for the sake of contradiction, suppose that $\sum v_i = 0$ with $v_i \in V_i$ and w.l.o.g. $v_1 \ne 0$. But then $$0 = A_1 (v_1 + \dots + v_k) = v_1 + A_1 v_2 + \dots + A_1 v_k = v_1 + A_1 A_2 v_2 + \dots + A_1 A_k v_k = v_1$$ since $A_1 A_i = 0$ for all $i$.

Remark. Here is a different argument for $AB = BA = 0$, without eigenvectors: multiplying by $A$ and using its idempotence and the super-commutativity, we have $$-BA = AB = A^2 B = AAB = -ABA = BAA = BA^2 = BA$$ thus $BA = 0$.