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SEEMOUS 2012 Problem 3


a) Prove that if k is an even positive integer and A is a real symmetric n \times n matrix such that (\text{Tr}\, (A^k))^{k+1}=(\text{Tr}\, (A^{k+1}))^k then A^n=\text{Tr}\, (A)A^{n-1}.

b) Does this assertion from a) also hold for odd positive integers k?

a) The equality given is equivalent to

\displaystyle (\lambda_1^k+...+\lambda_n^k)^{k+1}=(\lambda_1^{k+1}+...+\lambda_n^{k+1})^k, where \lambda_i are the eigenvalues of A, and they are real numbers since A is a real symmetric matrix. But we know that if a_1,...,a_n\geq 0 and t>1 then (a_1+...+a_n)^t-a_1^t-...-a_n^t\geq 0, with equality if all but one of a_i is equal to zero. To prove this, take \displaystyle f(a_n)= (a_1+...+a_n)^t- a_1^t - ... - a_n^t . Then f^\prime(a_n)=t(a_1+...+a_n)^{t-1}-ta_n^{t-1}\geq 0 and equality holds only for a_n=0. Therefore, the function f(a_n) is strictly increasing, and we have f(a_n)\geq f(0) with equality if and only if a_n=0. But notice that taking a_n=0 reduces the problem to the one for n-1. Doing this inductively, we notice that the inequality is true, and the equality holds if and only if all but one a_i are zero.

In our problem we have

\displaystyle \displaystyle (\lambda_1^k+...+\lambda_n^k)^{k+1}\geq (|\lambda_1|^k+...+|\lambda_n|^k)^{k+1}=

\displaystyle \displaystyle =[(|\lambda_1|^k+...+|\lambda_n|^k)^k]^{\frac{k+1}{k}} \geq \left( (|\lambda_1|^k)^{\frac{k+1}{k}}+...+(|\lambda_n|^k)^{\frac{k+1}{k}}\right)^k\geq

\geq \left(\lambda_1^{k+1}+...+\lambda_n^{k+1} \right)^k.

Since we have equality, all but maybe one of the eigenvalues of A are zero. Since A is diagonalizable, it follows immediately that A^n=\text{Tr}\, (A)A^{n-1}

Second proof, much easier than the first one, given by Filip Laurian, University of Bucharest. Suppose \lambda_1 has the greatest modulus among all \lambda_k. Then we have

\displaystyle (\lambda_1^k+...+\lambda_n^k)^{k+1}=(\lambda_1^{k+1}+...+\lambda_n^{k+1})^k\leq

\displaystyle (|\lambda_1|^{k+1}+...+|\lambda_n|^{k+1})^k\leq |\lambda_1|^k (\lambda_1^k+...+\lambda_n^k)^{k}

If all \lambda_i are zero, then we are done. Else, we can simplify the inequality above and get

\displaystyle |\lambda_1|^k+...+|\lambda_n|^k\leq |\lambda_1|^k

From here it follows immediately that all but, maybe \lambda_1 are zero, and the desired equality is immediate.

b) Take k=3 and A be 16 \times 16 with first 8 diagonal entries equal to (-1)^k and the other 8 equal to 1. Then \text{Tr} \, (A^3)=8 and \text{Tr} \, (A^4)=16. Since 8^4=16^3 it follows that A satisfies the given condition. But A^{16}=I and \text{Tr}\, (A) A^{15}=4A. Therefore, the conclusion does not hold.

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