About adding a negative definite rank-1 matrix to a symmetric matrix If $B$ is a symmetric matrix then how do its eigenvalues compare to the eigenvalues of $B - vv^T$? ( where $v$ is a vector of the same dimension as $B$)

I guess that the eigenvalues of $B - vv^T$ will downward interlace the eigenvalues of $B$. Like $\lambda_{min} (B - vv^T) \leq \lambda_{min} (B)$. 
 A: As suggested in the comments, this follows from the variational characterization of eigenvalues of symmetric matrices. For example, $\lambda_{\min}(A) =  \min_{x\cdot x =1} x\cdot A x$.
Let $y$ minimize $y\cdot B y$, then:
$$\begin{aligned}
\lambda_{\text{min}}(B-vv^T) &= \min x\cdot(B-vv^T)x\\
& \le y\cdot(B-vv^T)y\\
& = y\cdot By-(v\cdot y)^2\\
& = \lambda_{\text{min}} (B) - (v\cdot y)^2 \le \lambda_{\text{min}} (B)
\end{aligned}$$
A: I'm not sure if this is what you're looking for, but you could apply standard rank one theory to your problem. First of all, I want to assume that $v$ is cyclic for $B$; if this is not the case, then I can restrict attention to the reducing subspace $V$ that is spanned by the $B^nv$, $n\ge 0$. On $V^{\perp}$, the matrices $B$ and $A=B-vv^*$ agree.
Under this extra assumption, $v$ is then also cyclic for $A$.
Write $F(z) = v^*(B-z)^{-1}v$, $G(z)=v^*(A-z)^{-1}v$ for the matrix elements of the resolvents. From the resolvent identity
$$
(A-z)^{-1} - (B-z)^{-1} = (A-z)^{-1} vv^*(B-z)^{-1}
$$
I obtain that
$$
G(z) = \frac{F(z)}{1-F(z)} .
$$
Thus the eigenvalues of $A$ are the points where $F=1$. This reproves the interlacing property (because $F$ increases from $-\infty$ to $\infty$ between two consecutive eigenvalues of $B$) and gives somewhat more explicit information.
A: It seems to be "Cauchy's Interlacing Theorem" -- see Lemma 3.4 here:
http://arxiv.org/pdf/1408.4421v1.pdf
A: Since $B-(B-vv^T)$ is positive semidefinite, the $j$-th largest eigenvalue of $B$ is no less then the $j$-th largest eigenvalue of $B-vv^T$. This is a pretty known result. The proof is a quick use of min-max theorem; see  https://en.wikipedia.org/wiki/Min-max_theorem 
A: An interlacing result is in Theorem 4.3.4 of Horn, Johnson, Matrix Analysis (first edition - sorry, I don't have a copy of the second): for each Hermitian $B$,
$$
\lambda_k(B) \leq \lambda_{k+1}(B\pm vv^*) \leq \lambda_{k+2}(B), \quad k=1,2,\dots,n-2.
$$
