# International Mathematics Competition for University Students 2020

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IMC 2021
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## IMC2020: Day 1, Problem 2

Problem 2. Let $\displaystyle A$ and $\displaystyle B$ be $\displaystyle n\times n$ real matrices such that

$\displaystyle \textrm{rk}(AB-BA+I)=1$

where $\displaystyle I$ is the $\displaystyle n\times n$ identity matrix.

Prove that

$\displaystyle \tr(ABAB)-\tr(A^2B^2)=\frac12 n(n-1).$

($\displaystyle \textrm{rk}(M)$ denotes the rank of matrix $\displaystyle M$, i.e., the maximum number of linearly independent columns in $\displaystyle M$. $\displaystyle \tr(M)$ denotes the trace of $\displaystyle M$, that is the sum of diagonal elements in $\displaystyle M$.)

Rustam Turdibaev, V. I. Romanovskiy Institute of Mathematics

Solution. Let $\displaystyle X=AB-BA$. The first important observation is that

$\displaystyle \tr(X^2)=\tr(ABAB-ABBA-BAAB+BABA)=2\tr(ABAB)-2\tr(A^2B^2)$

using that the trace is cyclic. So we need to prove that $\displaystyle \tr(X^2)=n(n-1)$.

By assumption, $\displaystyle X+I$ has rank one, so we can write $\displaystyle X+I=v^tw$ for two vectors $\displaystyle v,w$. So

$\displaystyle X^2=(v^tw-I)^2=I-2v^tw+v^twv^tw=I+(wv^t-2)v^tw.$

Now by definition of $\displaystyle X$ we have $\displaystyle \tr(X)=0$ and hence $\displaystyle wv^t=\tr(wv^t)=\tr(v^tw)=n$ so that indeed

$\displaystyle \tr(X^2)=n+(n-2)n=n(n-1).$

An alternative way to use the rank one condition is via eigenvalues: Since $\displaystyle X+I$ has rank one, it has eigenvalue $\displaystyle 0$ with multiplicity $\displaystyle n-1$. So $\displaystyle X$ has eigenvalue $\displaystyle -1$ with multiplicity $\displaystyle n-1$. Since $\displaystyle \tr(X)=0$ the remaining eigenvalue of $\displaystyle X$ must be $\displaystyle n-1$. Hence

$\displaystyle \tr(X^2)=(n-1)^2+(n-1) \cdot 1^2=n(n-1).$

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