# trace and spectral radius inequality to symmetric positive definite matrix

If $\rho(A)$ is o spectral radius of $A$ and $tr(A)$ denotes the trace of $A$. The question is the following:

Let be $P,Q \in M_n(R)$ symmetric semidefinite positive matrix, then

$$tr(PQ)\le tr(P)\rho(Q)$$

I know that is necessary to use that the Frobenius and spectral norms are sub-multiplicative norms, and express the Frobenius norm how a sum of eigenvalues, but I have not idea yet. Help, please

## 1 Answer

Hint: $$\operatorname{tr}PQ=\operatorname{tr}P^{1/2}P^{1/2}Q^{1/2}Q^{1/2}= \operatorname{tr}P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=\|Q^{1/2}P^{1/2}\|_F^2\le\ldots$$ Another way: $$Q\le\rho(Q)\cdot I\implies P^{1/2}QP^{1/2}\le P^{1/2}\rho(Q)P^{1/2}=\rho(Q)\cdot P\implies\ldots$$

• I do not understand, yet, sorry. How do you define Q<=\rho(Q)I for Q and I matrices? in the second line – Cristian Camilo Espitia Morill Aug 14 '18 at 16:57
• @CristianCamiloEspitiaMorill Sorry for not mentioning that. For symmetric matrices $A\ge B$ often means that $A-B$ is positive semidefinite. – A.Γ. Aug 14 '18 at 17:44