Home > Algebra, Higher Algebra, Olympiad > Zero trace implies zero product

Zero trace implies zero product

A real positive definite matrix is a symmetric matrix whose eigenvalues are all nonnegative. Prove that if P,Q are real positive semidefinite n \times n matrices with tr(PQ)=0, then PQ=0.

Solution: An elegant solution to this problem can be given using the Cholesky decomposition of a symmetric positive definite real matrix. This means that for every positive semidefinite real matrix A there exists a real matrix L such that A=LL^t. We apply this for our matrices P,Q, and therefore P=XX^t,\ Q=YY^t, with X,Y real matrices. Now we have 0=tr(PQ)=tr(XX^tYY^t)=tr(Y^tXX^tY)=tr((Y^tX)(Y^tX)^t)=tr(AA^t) where A=Y^tX. But we know that tr(AA^t)=\sum_{i,j}a_{ij}^2, and this being equal to zero is equivalent to A=0. Then A^t=X^tY=0, and finally PQ=XX^tYY^t=0.

We can find using the argument presented above that tr(PQ) \geq 0, for all positive semidefinite real matrices P,Q.

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