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I'm starting by a simple remark: if $A$ is a $n\times n$ matrix and $\{\lambda_1,\ldots,\lambda_k\}$ are its eigenvalues, then the eigenvalues of matrix $I+A$ (where $I$ is the identity matrix) are $\{\lambda_1+1,\ldots,\lambda_k+1\}$. Moreover, if $\alpha\in\mathbb R$, the eigenvalues of $\alpha I+A$ are $\{\lambda_1+\alpha,\ldots,\lambda_k+\alpha\}$.

Are there more general results for this topic? Specifically, if $A$ is a $n\times n$ matrix and $\{\lambda_1,\ldots,\lambda_k\}$ are its eigenvalues, what are the eigenvalues of $A+D$ (where $D$ is a diagonal matrix)?

Edit (First): I was wondering if the solution is known in the case where the sum of the elements of every row of $A$ is $0$ and all the entries of $D$ is between $0$ and $1$.

Edit (Second): I was wondering if the solution is known when the sum of the elements of every row of $A$ is $0$ and $D=\operatorname{diag}(1, 0,\dots,0)$.

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  • $\begingroup$ See mathoverflow.net/q/90861/4612 $\endgroup$ – vadim123 Jul 26 '15 at 19:51
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    $\begingroup$ If $A$ and $D$ commute (multiplicatively!) then each eigenvalue of $A+D$ will be given by one of the eigenvalues of $A$ plus one of $D$. $\endgroup$ – David Simmons Jul 26 '15 at 20:04
  • $\begingroup$ If they are hermitian, you can use things like the (dual) Weyl or (dual) Lidskii inequalities to bound the eigenvalues. $\endgroup$ – David Simmons Jul 26 '15 at 20:06
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Two diagonalizable matrices $A$ and $B$ will commute if and only if they are simultaneously diagonalizable; i.e., we can write $$A = P^{-1} D_A P$$ and $$B=P^{-1} D_B P$$ where $D_A$ and $D_B$ are diagonal, and the elements of $D_A$ and $D_B$ represent the eigenvalues of $A$ and $B$, respectively.

Pointed out by Ian in the comments, a nondiagonalizable matrix commutes with some matrices, in particular itself.

Now, observe the following for simultaneously diagonalizable $A$ and $B$: $$A+B = P^{-1}D_AP + P^{-1}D_BP=P^{-1}(D_A+D_B)P.$$ Consequently, the eigenvalues of $A+B$ are given by the elements on the diagonal of $D_A+D_B$.

Note: considering you example, $A$ and $B=\alpha I$, $A$ and $B$ are simultaneously diagonalizable.

Extra note: It is important to recognize that for the general case that I have mentioned, there is some ambiguity as to precisely what the eigenvalues are. More precisely, the equality $$\lambda_{A+B,i} = \lambda_{A,i}+\lambda_{B,i}$$ does not hold in general. However, given a $\lambda_{A+B,i}$, it is always possible to find a $\lambda_{A,j}$ and $\lambda_{B,k}$ such that $$\lambda_{A+B,i} = \lambda_{A,j}+\lambda_{B,k}.$$ I'm not sure if anyone else can elaborate on this.

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    $\begingroup$ Pedantic point: two diagonalizable matrices commute if and only if they are simultaneously diagonalizable. A nondiagonalizable matrix commutes with some matrices, in particular itself. $\endgroup$ – Ian Jul 26 '15 at 20:44
  • $\begingroup$ Yes, I'll make the edit. $\endgroup$ – David Simmons Jul 26 '15 at 21:01
  • $\begingroup$ Thank you, I voted your answer as "useful", but my matrices don't commute. $\endgroup$ – Mark Jul 27 '15 at 4:57
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There is no precise answer to your question. Indeed, let $A,B\in M_n(\mathbb{C}),spectrum(A)=(a_i),spectrum(B)=(b_i),spectrum(A+B)=(c_i)$. Then the possible values of $(a_i,b_i,c_i)$ are dense in the set $\sum_i a_i+\sum_i b_i=\sum_i c_i$. Since we reason by density, we may assume that $B$ is the diagonal $D=diag((b_i))$. If $A,B$ are real, then we must add the fact that the eigenvalues of the considered matrices are real or pairwise conjugate. Now I do not think that your condition $A.vect(1)=0$ adds some useful information.

EDIT. About the David's post. Let $A,B\in M_n(\mathbb{C}),spectrum(A)=(a_i),spectrum(B)=(b_i)$; if $AB=BA$, then there is a permutation of the $(b_i)$ s.t., for every $u,v\in\mathbb{C}$, $spectrum(uA+vB)=(ua_i+vb_i)$.

Proof. $A,B$ are simultaneously triangularizable.

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I'm not sure if what it is motivating your question is the same that I have in mind, but at least here is an answer for your question for the case when the matrix $A$ is symmetric, positive semidefinite, satisfies $\text{kernel}(A)=\text{span}\{\textbf{1}\}$, and $D=\text{diag}([0,0,..,0,d_i,0...,0])$ with $d_i>0$, which is similar to what you have.

Since $A$ and $D$ are symmetric, the smallest eigenvalue of $(A+D)$ is given by

\begin{equation} \lambda_{min}=\min_{z:z^\top z=1} z^\top(A+D)z=\min_{z:z^\top z=1} z^\top Az+z^\top Dz \end{equation}

Since $A$ and $D$ are P.S.D, we have that $z^\top Az\geq0$ and $z^\top Dz\geq0$ for all $z$. Moreover, since $kernel(A)=span\{\mathbf{1}\}$, the term $z^\top Az\geq0$ can only be zero if $z\in\text{span}\{\mathbf{1}\}$, but for any $z\in\text{span}\{\mathbf{1}\}$ we have that $z^\top Dz>0$ since $D$ is diagonal with at least one positive entry. Therefore we have that $\lambda_{min}>0$.

The same reasoning can be used to find an upper bound on $\lambda_{max}$.

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