= Rank-one downdate along an eigenvector
{title2=$(A-\rho bb^T)b=(\lambda-\rho)b,\quad (A-\rho bb^T)|_{b^\perp}=A|_{b^\perp}$}
Let $A$ be a real symmetric <positive-definite matrix>, and let $b$ be a unit <eigenvector> with <eigenvalue> $\lambda>0$. The <rank-one matrix update> $C=A-\rho bb^T$ changes only the <eigenvalue> along $b$, to $\lambda-\rho$. <Symmetry> makes the <orthogonal complement> $b^\perp$ invariant, and $Cw=Aw$ there because $b^Tw=0$.
In detail, write $x=tb+w$ with $w\perp b$. Then
$$
x^TCx=(\lambda-\rho)t^2+w^TAw.
$$
Consequently $C$ is a <positive-definite matrix> for $\rho<\lambda$, singular for $\rho=\lambda$, and has exactly one negative <eigenvalue> for $\rho>\lambda$. In dimension at least two the last case is indefinite. In dimension one it is <negative definite> instead: there is no nonzero orthogonal-complement direction. The distinction matters when classifying a <stationary point> of a <quadratic form>.
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