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Assume that A, B are positive n by n matrices and the rank of B is 1, B=xx*. If the eigenvalues of A are a_1≥a_2≥...≥a_n, and x is not the eigenvector of A, then there are d_i≥0 such that eigenvalue of A+B are a_1+d_1, a_2+d_2,...,a_n+d_n. Is it true? Are d_is non-negative?

My e-mail is kanhemath@yahoo.com.cn If you any point, please let me know.

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Yours He

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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. –  Noah Stein Mar 29 '12 at 14:54

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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:

http://books.google.ie/books?id=PlYQN0ypTwEC&pg=PA198&dq=interlacing+horn+johnson&hl=en&sa=X&ei=2HZ0T_DrIZOAhQeeqKSmBQ&redir_esc=y#v=onepage&q&f=false

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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That is a good job! Thanks! –  Kan He Mar 30 '12 at 10:48

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