Consider an $n\times n$ matrix $A$ that is positive semidefinite and has rank $n-1$, so there exists exactly one eigenvector $v$ such that $Av=0$. Let now $B$ be a symmetric matrix such that $v^TBv=0$. I'd like to argue that, if the norm of $B$ is small enough, $A+B$ is always positive semidefinite.
Perturbation of positive semidefinite matrix
linear algebramatricespositive-semidefinitesymmetric matrices
Best Answer
The condition that $v^TBv$ is insufficient; we require the stronger condition that $Bv = 0$. As a counterexample, take $n = 2$ and $$ v = \pmatrix{1\\0}, \quad A = \pmatrix{0 & 0\\0&1}, \quad B = \pmatrix{0&t\\t&0}, \quad t > 0. $$ For any $t > 0$, we find that $A + B$ fails to be positive semidefinite.
If $Bv = 0$, then we can argue that $A + B$ is positive semidefinite as follows. Let $\{v_1,v_2,\dots,v_n\}$ be an orthonormal basis with $v_1 = v$. Let $Q$ denote the orthogonal matrix whose columns are $v_1,\dots,v_n$, and break it into the block-matrix $Q = [v\ \ Q_\perp]$. We can replace $A$ and $B$ with the similar matrices $$ A_0 = Q^TAQ = \pmatrix{v^TAv & v^TAQ_\perp \\ Q_\perp^T Av & Q_\perp^TAQ_\perp}, \quad B_0 = Q^TAQ = \pmatrix{v^TBv & v^TBQ_\perp \\ Q_\perp^T Bv & Q_\perp^TBQ_\perp}. $$ We note that $Bv = 0$ and $v^TB = (Bv)^T = 0$ and similarly $Av = 0$ and $v^TA = 0$. Thus, we have $$ Q^T(A + B)Q = A_0 + B_0 = \pmatrix{0& 0\\0 & Q_\perp^TAQ_\perp + Q_\perp^TBQ_\perp}, $$ which is indeed positive semidefinite if the norm of $B$ (and hence $Q_\perp^TBQ_\perp$) is sufficiently small.
With this decomposition, we also see that if $v^TBv = 0$, then $A + B$ will be positive semidefinite only if $Bv = 0$. Otherwise, $Q_\perp^Tv$ will be non-zero, which means that $Q^T(A + B)Q$ will have a $2 \times 2$ principal submatrix that fails to be positive semidefinite.