## SEEMOUS 2012 Problem 3

a) Prove that if is an even positive integer and is a real symmetric matrix such that then .

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

a) The equality given is equivalent to

, where are the eigenvalues of , and they are real numbers since is a real symmetric matrix. But we know that if and then , with equality if all but one of is equal to zero. To prove this, take . Then and equality holds only for . Therefore, the function is strictly increasing, and we have with equality if and only if . But notice that taking reduces the problem to the one for . Doing this inductively, we notice that the inequality is true, and the equality holds if and only if all but one are zero.

In our problem we have

Since we have equality, all but maybe one of the eigenvalues of are zero. Since is diagonalizable, it follows immediately that

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

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

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

b) Take and be with first diagonal entries equal to and the other equal to . Then and . Since it follows that satisfies the given condition. But and . Therefore, the conclusion does not hold.