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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?

Thanks in advanced

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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 $\Lambda_B$.

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would you say a reference for this. " two commutative matrices are diagonalizable", thanks – Vahid Aug 20 '13 at 5:37
Commuting matrices over an algebraically closed field are simultaneously triangularizable!! but not diagonalizable!!? – Vahid Aug 25 '13 at 5:33
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
do you have any idea for operators which have infinite eigenvalues? – Vahid Aug 25 '13 at 5:36

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.

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