IMC 2021 · Day 1 · P1

Hint: (a) Multiply the equation by some power of $A$ from left and another power of $A$ from right. (b) Substitute repeatedly $X = A - AX - XA^2$.

First suppose that some matrix $X$ satisfies the equation. We can obtain new equations if we multiply the given equation by some power of $A$ from left and another power of $A$ from right. For example, $$A^2 (X + AX + XA^2) A^2 = A^2 X A^2 + A^3 \cdot X A^2 + A^2 X A \cdot A^3 = A^2 X A^2.$$ The right-hand side is $A^2 \cdot A \cdot A^2 = A^3 \cdot A^2 = 0$, so $$A^2 X A^2 = A^2 (X + AX + XA^2) A^2 = A^5 = 0. \quad \text{Similarly,}$$ $$\begin{gather*} A^2 X = A^2 (X + AX + XA^2) = A^3 = 0 \\ A X A = A (X + AX + XA^2) A = A^3 = 0 \\ X A^2 = (X + AX + XA^2) A^2 = A^3 = 0 \\ A X = A (X + AX + XA^2) = A^2. \quad \text{Finally} \\ X = A - AX - XA^2 = A - A^2. \end{gather*}$$ Hence, no matrix other than $A - A^2$ can satisfy the equation.

Note that the argument above does not prove that the matrix $X = A - A^2$ satisfies the equation, because the steps cannot be done in reverse order. That must be verified separately. Indeed, $$X + AX + XA^2 = (A - A^2) + A (A - A^2) + (A - A^2) A^2 = A - A^4 = A.$$ Hence, $X = A - A^2$ is the unique solution of the equation.

Remark. By multiplying the equation by $A^n$ from left and by $A^k$ from right we can get 9 different equations: $$\begin{gather*} X + AX + XA^2 = A \qquad XA + AXA = A^2 \qquad XA^2 + AXA^2 = 0 \\ AX + A^2 X + AXA^2 = A^2 \qquad AXA + A^2 XA = 0 \qquad AXA^2 + A^2 XA^2 = 0 \\ A^2 X + A^2 XA^2 = 0 \qquad A^2 XA = 0 \qquad A^2 XA^2 = 0 \end{gather*}$$ These formulas provide a system of linear equations for the nine matrices $X$, $AX$, $A^2X$, $XA$, $AXA$, $A^2XA$, $XA^2$, $AXA^2$ and $A^2XA^2$.

We use a different approach to express $X$ in terms of $A$. If some matrix $X$ satisfies the equation then $$X = A - AX - XA^2.$$ Let us substitute this identity in the right-hand side repeatedly until $X$ cancels out everywhere.

Notice that by the condition $A^3 = 0$ we have $A^3 = A^4 = A^5 = A^3 X = X A^4 = A X A^4 = A^3 X A^2 = 0$, so $$\begin{align*} X &= A - AX - XA^2 \\ &= A - A (A - AX - XA^2) - (A - AX - XA^2) A^2 \\ &= A - (A^2 - A^2 X - A X A^2) - (A^3 - A X A^2 - X A^4) \\ &= A - A^2 + A^2 X + 2 A X A^2 \\ &= A - A^2 + A^2 (A - AX - XA^2) + 2 A (A - AX - XA^2) A^2 \\ &= A - A^2 + (A^3 - A^3 X - A^2 X A^2) + 2 (A^4 - A^2 X A^2 - A X A^4) \\ &= A - A^2 - 3 A^2 X A^2 \\ &= A - A^2 - 3 A^2 (A - AX - XA^2) A^2 \\ &= A - A^2 - 3 (A^5 - A^3 X A^2 - A^2 X A^4) \\ &= A - A^2. \end{align*}$$ To complete the solution, we have to verify that $X = A - A^2$ is indeed a solution. This step is the same as in Solution 1.

Let $B = I - A + A^2$ so that $B$ is the inverse of $I + A$. Multiplying by $B$ from the left, the equation is equivalent to $$\tag{1} X + B X A^2 = B A.$$ Now assume $X$ satisfies the equation. Multiplying by $A^2$ from the right and using $A^3 = 0$ we get $X A^2 = 0$. Hence the equation simplifies to $X = BA = A - A^2$.

On the other hand, $X = BA$ obviously satisfies (1).