Change in kernel after rank-1 update

Question

Let $B$ be an $n\times n$ symmetric and positive semidefinite matrix and $v\in \mathbb{R}^n$ a vector. Consider the positive semidefinite matrix $A=B+vv^T$. It is clear that $\mathrm{ker}(B)\cap (v)^\perp \subseteq \mathrm{ker}(A)$. Is the opposite also true? That is I want to know whether $\mathrm{ker}(A) = \mathrm{ker}(B)\cap (v)^\perp$. If this is true, then is the condition "positive semidefinite" actually necessary?

A matrix $X$ is positive semidefinite if $u^TXu \geq 0$ for every $u\in\mathbb{R}^n$.

Answer 1

Let $A=B+C$ with $B$ and $C$ symmetric positive semi-definite, then $\ker(A)=\ker(B)∩\ker(C)$.

Proof:

• "⊆": $Ax=0 ⟹ x^⊤ Bx+ x^⊤ Cx = 0 ⟹ x^⊤Bx=0$ and $x^⊤ C x=0 ⟹ Bx=0$ and $Cx=0$. The last 2 steps follow from the positive semi-definiteness of $B$ and $C$.

• "⊇": $Bx=0$ and $Cx=0 ⟹ Bx+Cx=0 ⟹ Ax=0$.

The positive semi-definiteness is necessary, even if $C=vv^⊤$, because otherwise consider $B=-C$.