Can a given symmetric matrix be written as a linear combination of the identity and a rank-$1$ matrix?

Question

Given an invertible symmetric matrix $M$, is there a way to determine whether it can be written as $$M = a v v^T + b \mathrm{I}$$ for some scalars $a,b \neq 0$ and a vector $v$? I'm given matrix $M$ and want to find $a, b$ and $v$.

Answer 1

If $M$ is an $n\times n$ matrix of the form $M=avv^T+bI$, then $M$ is clearly symmetric and we may always suppose that $v^Tv=1$, by replacing $v$ with $v'=(v^Tv)^{-1/2}v$, and $a$ with $a'= (v^Tv)a$.

Assuming henceforth that $v^Tv=1$, observe that $$ Mv=avv^Tv+bv = (a+b)v, $$ and that, for every vector $u$ perpendicular to $v$, we have $$ Mu=avv^Tu+bu = bu. $$ Therefore the spectrum of $M$ coincides with $\{a+b,b\}$, where the multiplicity of $a+b$ is $1$ (provided $a\neq 0$), and the multiplicity of $b$ is $n-1$.

Conversely, every symmetric matrix whose spectrum consists of exactly two points, say $\lambda _1$ and $\lambda _2$, with multiplicities 1 and $n-1$, respectively, may be written as $$ M=avv^T+bI, $$ where $b$ is the spectral value with multiplicity $n-1$, namely $b=\lambda _2$, and $a=\lambda _1-\lambda _2$. Finally, $v$ is any unit eigenvector associated to $\lambda _1$.