Rank-one downdate along an eigenvector ($(A-\rho b{b}^T)b=(\lambda -\rho )b,(A-\rho b{b}^T){|}_{b^{\perp }}=A{|}_{b^{\perp }}$)

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 $C=A-\rho b{b}^T$ changes only the eigenvalue along $b$, to $\lambda -\rho$. Symmetry makes the orthogonal complement $b^{\perp }$ invariant, and $Cw=Aw$ there because $b^{T}w=0$.

In detail, write $x=tb+w$ with $w\perp b$. ThenConsequently $C$ 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.