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. $\endgroup$ – Gerry Myerson Jul 27 '15 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. $\endgroup$ – Gerardo Arizmendi Jul 27 '15 at 8:52
  • $\begingroup$ @Gerardo, note that one of the eigenvalues of $A$ is zero. $\endgroup$ – Gerry Myerson Jul 27 '15 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 '15 at 12:48
  • 1
    $\begingroup$ 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 '15 at 15:11

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 '15 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 '15 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 '15 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 '15 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.