Assume that $A, B$ are positive $n \times n$ matrices and that $B$ is rank-$1$, i.e., $B=xx^*$. If the eigenvalues of $A$ are $a_1 \geq a_2 \geq \cdots \geq a_n$, and $x$ is not an eigenvector of $A$, then there are $d_i \geq 0$ such that eigenvalues of $A+B$ are $a_1+d_1, a_2+d_2,\dots,a_n+d_n$. Is it true? Are the $d_i$s non-negative?


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    $\begingroup$ I think you need to provide a bit more information. Do your matrices have real or complex entries? By positive, do you mean positive semidefinite? Also, you seem to be assuming the matrices are symmetric? Please edit your post to include this information and check that your question is complete before posting. $\endgroup$
    – Noah Stein
    Mar 29, 2012 at 14:54

1 Answer 1


Yes, provided you assume Hermitianity. Then, even more is true.

Take A to be Hermitian (or real symmetric, if you like) matrix. As for B, it can be any positive semidefinite matrix (including your rank 1 case and without regard to the eigenvectors of A). Then your assertion follows from Weyl's Theorem about the eigenvalues of the sum of Hermitian matrices. This is actually stated as Problem 1 on page 198 of the Horn & Johnson Matrix Theory.

Here's a Google Books link to it:


Since B is positive semidefinite, $\lambda_{1}(B) \geq 0$.

As you can see there, you can even bound the $d_{i}$'s from above.


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