# eigenvalues of two positive commutative matrices

Let $A$ and $B$ be two positive commutative matrices. I am going to prove

$$\lambda_{j}(A+B)\leq \lambda_{j}(A)+\lambda_{j}(B)$$

for $j=1,2,\ldots n$, where $\lambda_{j}$ are eigenvalues of matrix and these eigenvalues are in a decreasing order as follows $$\lambda_{1}\geq\lambda_{2}\geq\ldots\geq\lambda_{n}$$ I heard that $\lambda_{max}$ or $\lambda_{1}$ is correct for the first inequality, I mean $\lambda_{1}(A+B)\leq \lambda_{1}(A)+\lambda_{1}(B)$ ?

Can we take out the max eigenvalue and say the similar thing for others?

Is it true for operators which may have infinite eigenvalues?

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Hint: By positive commutative matrices, I suppose you mean they are positive definite and commute with each other. The first implies that all their eigenvalues are positive and the second one implies that they are simultaneously diagonalizable, thus \begin{align} \mathbf{A}=\mathbf{U}\Lambda_A\mathbf{U}^H~~~~\mathbf{B}=\mathbf{U}\Lambda_B\mathbf{U}^H \end{align} where $\Lambda_i$ is the diagonal eigenvalue matrix for $i=A,B$. Thus \begin{align} \mathbf{A}+\mathbf{B}=\mathbf{U}(\Lambda_A+\Lambda_B)\mathbf{U}^H \end{align} Now think about the order of eigenvalues in $\Lambda_A$ and $\Lambda_B$.
we can say if $A$ and $B$ are diagonalizable and commuting together then they are simultaneously diagonalizable.Am I right? –  Vahid Aug 25 '13 at 5:35
It is not correct in general, Let $A=[1,0;0,0]$ and $B=[0,0;0,1]$ then $\lambda_{2}$ does not satisfy the inequality.