Finding a symmetric matrix E such that the approx eigenpair for A is the exact eigenpair for A+E

Question

Let $A$ be a real $N\times N$ matrix. If $(x^TAx, x)$ is an approximate eigenpair for $A$ ($x$ a normalised vector) then find a symmetric $E$ such that $(x^TAx, x)$ is an exact eigenpair for $A+E$.

Just can't figure out how to go about this. Thanks for any insight,

Beans x

Answer 1

Let $r:=Ax-\rho x$ with $\rho:=x^TAx$ and $x^Tx=1$. If $E:=-(xr^T+rx^T)$ (clearly $E$ is symmetric), then $$(A+E)x=Ax-(r^Tx)x-(x^Tx)r=Ax-(r^Tx)x-r.$$ We have $r^Tx=(Ax-\rho x)^Tx=x^TAx-\rho x^Tx=0$ so $$ (A+E)x=Ax-r=\rho x, $$ that is, $(\rho,x)$ is an eigen-pair of $A+E$.