# Perturbation of positive semidefinite matrix

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.

• This is correct. This can be proven using standard perurbation results. See the book by Kato for instance.
– KBS
Jun 6 at 20:18
• Thank you @KBS. Can you please tell me which result you are referring to in Kato's book? Jun 6 at 20:50

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.

Assuming we require non-zero $$B$$, then the claim is False. Consider the case of $$n=2$$ with
$$A=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$$ and $$B=\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}$$.
For any $$\lambda \in \mathbb R-\big\{0\big\}$$

$$A+\lambda B = \begin{bmatrix}1 +\lambda & \lambda \\ \lambda & 0\end{bmatrix}$$
which is a real symmetric matrix that has a zero on its second diagonal component but the entire row fails to be zero hence is indefinite. Alternatively just check that the determinant is $$\lambda^2 \lt 0$$