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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.