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Question:

Let $A$ and $B$ be two real $n×n$ matrices which commute and let $a_1, a_2,..., a_n$ (not necessarily distinct) be the eigenvalues of $A$ and $b_1 ,...,b_n$ are eigenvalues (not necessarily distinct) of $B$.

(1) Are $a_1 + b_1,...,a_n +b_n$ the eigenvalues of $A+B$ ?

(2) If $a$ is eigenvalue of $A$ with algebraic multiplicity $k$ and $b$ is eigenvalue of $B$ with algebraic multiplicity $k$, then is $a+b$ an eigenvalue of $A+B$ with algebraic multiplicity $k$ ?

My attempt: I know that if $A$ & $B$ are commutating real matrices then they are simultaneously triangularizable & from this we can get, spectrum of $A+ B$ is contained in $\{a_1 + b_1 : a_1 ∈ σ(A) , b_1 ∈ σ(B) \}$ But then how can we get (1) and (2) from this?

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Just consider $A=\pmatrix{1&0\\ 0&-1}$ and the following two different cases: $B=A$ and $B=\pmatrix{-1&0\\ 0&1}$.

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  • $\begingroup$ So is that mean answer to about two questions is false? If so please can u tell me, when those are true? Is there is any condition? $\endgroup$ – Akash Patalwanshi Sep 24 '17 at 15:19
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    $\begingroup$ @AkashPatalwanshi Yes, both are false. $\endgroup$ – user1551 Sep 24 '17 at 15:20
  • $\begingroup$ So when they are true? if i consider $I$ as $101× 101$ identity matrix and J as $101× 101$ matrix with each entries $1$ then, eigenvalue of $I$ is $1$ with $A.M = 101$ and eigenvalues of $J $are $0 $with $A.M = 100$ and $1$with A.M = 1. Clearly $101- 1 = 100$ is eigenvalue of $J-I$ and $ 0-1 = -1 $ is eigenvalue of $J-I$. So in this case (1) is true. Why so? Is there is any reason? $\endgroup$ – Akash Patalwanshi Sep 24 '17 at 15:28
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    $\begingroup$ @AkashPatalwanshi If two matrices commute, they can be simultaneously triangularised over $\mathbb C$. So, if $a_i$ denotes the $i$-th diagonal entry in the aforementioned triangulation and similarly for $b_i$), then the eigenvalues of $A+B$ are indeed $a_i+b_i$. But (1) is not true in general because the order of eigenvalues is unspecified. Unless all eigenvalues in one matrix are identical, you can always scramble the eigenvalues to make (1) or (2) false. Your example works because all eignevalues of $A$ are equal to each other. $\endgroup$ – user1551 Sep 24 '17 at 15:36
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    $\begingroup$ @AkashPatalwanshi Yes, but note that if all eigenvalues of $A$ are the same, we must have $k=n$. In turn, the statement of (2) becomes "if all eigenvalues of $A$ are equal to $a$ and all eigenvalues of $B$ are equal to some $b$, then all eigenvalues of $A+B$ are equal to $a+b$". $\endgroup$ – user1551 Sep 24 '17 at 15:52

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