I know that given two matrices $A$ and $B$, estimating the eigenvalues of $A + B$ by the eigenvalues of $A$ and $B$ is generally a non-easy problem. In particular, there are some results for matrices that commute (multiplicatively!), hermitian matrices etc.

In this case $B=\operatorname{diag}(1, 0,\dots,0)$ and the sum of the elements of every row of $A$ is $0$; each eigenvalue of $A$ is non-negative. I was wondering if the solution is known in this case, at least if one can say something about the sign of eigenvalues of $A+B$.

Thanks in advance.

  • 1
    $\begingroup$ I doubt it. If $$A=\pmatrix{1&-1\cr1&-1\cr}$$ then one eigenvalues of $A+B$ is positive, and one is negative. $\endgroup$ Jul 27, 2015 at 6:21
  • $\begingroup$ You can say something, if the eigenvalues of A are big enough in norm, the eigenvalues of A+B will not change sign, cause they are continuous functions of the values of the matrix. $\endgroup$ Jul 27, 2015 at 8:52
  • $\begingroup$ @Gerardo, note that one of the eigenvalues of $A$ is zero. $\endgroup$ Jul 27, 2015 at 9:28
  • 2
    $\begingroup$ Have a look at: cs.vu.nl/~ran/LectureBerlijn2010.pdf --- basically, your problem is that of determining eigenvalues after a rank-one perturbation... $\endgroup$
    – Suvrit
    Jul 27, 2015 at 12:48
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    $\begingroup$ I think @GerryMyerson's comment answers the question, but if you want $A$ to have a nonzero eigenvalue you can take $A = \pmatrix{2&-2\cr 1&-1}$. Its eigenvalues are $0$ and $1$, but the eigenvalues of $A + B$ are $1 \pm \sqrt{2}$. $\endgroup$
    – Nik Weaver
    Jul 27, 2015 at 15:11

1 Answer 1


The question is to know if the $0$ eigenvalue of $A$ can become negative when we add $B$. We can write specific results only if $B$ is a small perturbation of $A$; it is easier to assume that $A$ is fixed and $B=diag(x,\cdots,0)$ with a small positive $x$.

Let $U$ be the matrix obtained from $A$ deleting its first column and row. Then $\phi(\lambda)=\det(A+B-\lambda I)=\det(U-\lambda I)x+\det(A-\lambda I)$; $\phi(0)=\det(U)x$ and $\phi'(0)=-tr(adj(U))x-tr(adj(A))$. Since $0$ is a simple eigenvalue of $A$, $tr(adj(A))\not=0$ and $0$ does not burst in $2$ conjugate eigenvalues of $A+B$. We assume that $\det(U)\not=0$.

EDIT. Here we assume that $0$ is a simple eigenvalue of $A$ and (consequently) its other eigenvalues are $>0$. Then $tr(adj(A))>0$. Thus the conclusion is as follows:

If $\det(U)<0$, then $A$ admits a negative eigenvalue; otherwise $0$ gives birth to a positive eigenvalue.

  • $\begingroup$ Thank you! What does mean that $A$ is great? $\endgroup$
    – Mark
    Aug 26, 2015 at 5:34
  • $\begingroup$ Yes, "great" is not the correct word. What I wanted to say is that $B$ must be a small perturbation of $A$. $\endgroup$
    – loup blanc
    Aug 26, 2015 at 10:09
  • $\begingroup$ Ok. Is this true even if $B$ is a "small" diagonal matrix (that is, with entry positive and near zero)? $\endgroup$
    – Mark
    Aug 27, 2015 at 12:38
  • $\begingroup$ If $B$ is a small matrix that is similar to $diag(x,0,\cdots,0)$, then we can prove a similar result. Now, if $B=diag(x_i)$ then the result does not work; for instance, if the $x_i$ are equal to $x>0$, then the eigenvalues of $A+B$ are $\geq x$ and, consequently, are always $>0$. $\endgroup$
    – loup blanc
    Aug 27, 2015 at 21:53

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