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Let us consider two arbitrary Hermitian square matrices $\mathbf{A,B}$ with the same dimension. Given $\mathbf{v}$ the eigenvector associated to the maximum eigenvalue of $\mathbf{A}$:

Are there any conditions appart from $\mathbf{A} = \mathbf{B}$ or $\mathbf{B} = \mathbf{v}\mathbf{v}^H$ so that it can be ensured that $\mathbf{B}$ has the same eigenvector associated to the maximum eigenvalue?

Note that, in Dan Shemesh, Common eigenvectors of two matrices, Linear Algebra and its Applications, Volume 62, November 1984, Pages 11-18, ISSN 0024-3795, http://dx.doi.org/10.1016/0024-3795(84)90085-5. an arbitrary eigenvector was considered. I cross check the derivation, and I do not see how to promote the maximum eigenvector constraint.

Thank you in advance

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    $\begingroup$ The question is unclear. Are they real matrices? (Otherwise what do you mean by "largest" --- largest in absolute value?) What kind of condition would you want? $\endgroup$
    – Nik Weaver
    Commented Jun 29, 2015 at 12:14
  • $\begingroup$ Thank you Nik, I add the condition of A and B being Hermitian. I look for a more 'exhaustive' condition rather than the ones I metion in the question. $\endgroup$
    – mikitov
    Commented Jun 29, 2015 at 12:19
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    $\begingroup$ One such condition is that (for normal matrices) $\| AB\| = \| A \| \cdot \|B\|$ (where obviously we take the $2-2$ norm). $\endgroup$
    – David Handelman
    Commented Jun 29, 2015 at 13:34
  • $\begingroup$ @DavidHandelman: maybe require $A$ and $B$ to be positive? I'm unsure whether the "maximum" eigenvalue is allowed to be negative ... $\endgroup$
    – Nik Weaver
    Commented Jun 29, 2015 at 15:19
  • $\begingroup$ $A$, $B$ are only required to be Hermitian. With this, the eigenvalues are always real, then there are no worries about the ordering. $\endgroup$
    – mikitov
    Commented Jun 29, 2015 at 15:56

1 Answer 1

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This is true iff the largest eigenvalue of $A+B$ is the sum of that of $A$ and that of $B$.

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  • $\begingroup$ Could you elaborate your response a little? $\endgroup$
    – mikitov
    Commented Jun 30, 2015 at 9:04
  • $\begingroup$ For an appropriate interpretation of the question, at least. The OP said $v$ is "the" eigenvector associated to the maximum eigenvalue of $A$, but if the eigenspace is not 1-dimensional you may need to replace $v$ with another eigenvector. $\endgroup$
    – Mark Meckes
    Commented Jun 30, 2015 at 11:01

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