# International Mathematics Competition for University Students 2021

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IMC 2021
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## IMC2021: Day 1, Problem 1

Problem 1. Let $\displaystyle A$ be a real $\displaystyle n\times n$ matrix such that $\displaystyle A^3=0$.

(a) Prove that there is a unique real $\displaystyle n\times n$ matrix $\displaystyle X$ that satisfies the equation

$\displaystyle X+AX+XA^2=A.$

(b) Express $\displaystyle X$ in terms of $\displaystyle A$.

Bekhzod Kurbonboev, Institute of Mathematics, Tashkent

Solution 1. First suppose that some matrix $\displaystyle X$ satisfies the equation. We can obtain new equations if we multiply the given equation by some power of $\displaystyle A$ from left and another power of $\displaystyle A$ from right. For example,

$\displaystyle A^2(X+AX+XA^2)A^2 = A^2XA^2+A^3\cdot XA^2+A^2XA\cdot A^3 = A^2XA^2.$

The right-hand side is $\displaystyle A^2\cdot A\cdot A^2=A^3\cdot A^2=0$, so

\begin{align*} A^2XA^2 &= A^2(X+AX+XA^2)A^2 = A^5 = 0. \qquad \text{Similarly,} \\ A^2X &= A^2(X+AX+XA^2) = A^3 = 0 \\ AXA &= A(X+AX+XA^2)A = A^3 = 0 \\ XA^2 &= (X+AX+XA^2)A^2 = A^3 = 0 \\ AX &= A(X+AX+XA^2)A = A^2. \qquad \text{Finally} \\ X &= A-AX-XA^2 = A-A^2. \end{align*}

Hence, no matrix other than $\displaystyle A-A^2$ can satisfy the equation.

Note that the argument above does not prove that the matrix $\displaystyle X=A-A^2$ satisfies the equation, because the steps cannot be done in reverse order. That must be verified separately. Indeed,

$\displaystyle X+AX+XA^2 =(A-A^2)+A(A-A^2)+(A-A^2)A^2 = A-A^4 = A.$

Hence, $\displaystyle X=A-A^2$ is the unique solution of the equation.

By multiplying the equation by $\displaystyle A^n$ from left and by $\displaystyle A^k$ from right we can get 9 different equations:

$\displaystyle \begin{matrix} X+AX+XA^2=A & XA+AXA=A^2 & XA^2+AXA^2=0 \\ AX+A^2X+AXA^2=A^2 & AXA+A^2XA=0 & AXA^2+A^2XA^2=0 \\ A^2X+A^2XA^2=0 & A^2XA=0 & A^2XA^2 =0 \\ \end{matrix}$

These formulas provide a system of linear equations for the nine matrices $\displaystyle X$, $\displaystyle AX$, $\displaystyle A^2X$, $\displaystyle XA$, $\displaystyle AXA$, $\displaystyle A^2XA$, $\displaystyle XA^2$, $\displaystyle AXA^2$ and $\displaystyle A^2XA^2$.

Solution 2. We use a different approach to express $\displaystyle X$ in terms of $\displaystyle A$. If some matrix $\displaystyle X$ satisfies the equation then

$\displaystyle X = A-AX-XA^2.$

Let us substitute this identity in the right-hand side repeatedly until $\displaystyle X$ cancels out everywhere. Notice that by the condition $\displaystyle A^3=0$ we have $\displaystyle A^3=A^4=A^5=A^3X=XA^4=AXA^4=A^3XA^2=0$, \displaystyle so \begin{align*} X &= A-AX-XA^2 \\ &= A-A(A-AX-XA^2)-(A-AX-XA^2)A^2 \\ &= A-(A^2-A^2X-AXA^2)-(A^3-AXA^2-XA^4) \\ &= A-A^2 +A^2X +2AXA^2 \\ &= A-A^2 +A^2(A-AX-XA^2) +2A(A-AX-XA^2)A^2 \\ &= A-A^2 +(A^3-A^3X-A^2XA^2) + 2(A^4-A^2XA^2-AXA^4) \\ &= A-A^2 -3A^2XA^2 \\ &= A-A^2 -3A^2(A-AX-XA^2)A^2 \\ &= A-A^2 -3(A^5-A^3XA^2-A^2XA^4) \\ &= A-A^2. \end{align*} To complete the solution, we have to verify thatX=A-A2$\displaystyle is indeed a solution. This step is the same as in Solution~1. Solution 3. Let \(\displaystyle B=I-A+A^2$ so that $\displaystyle B$ is the inverse of $\displaystyle I+A$. Multiplying by $\displaystyle B$ from the left, the equation is equivalent to

 $\displaystyle X+BXA^2=BA.$ $\displaystyle (1)$

Now assume $\displaystyle X$ satisfies the equation. Multiplying by $\displaystyle A^2$ from the right and using $\displaystyle A^3=0$ we get $\displaystyle XA^2=0$. Hence the equation simplifies to $\displaystyle X=BA=A-A^2$.

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

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