$A$ is symmetric and positive definite matrix of $n \times n$. $I$ is the Identity matrix of $n \times n$.
$a$ and $b$ are vectors of $n \times 1$.
$a.b \neq -1$ and $a.b \neq 0$
$a$ is not parallel to $Ab$
How do we show that $X-A$ is a rank $2$ matrix ?
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$ Hence each of the terms are having rank $1$. so the sum of all the terms can have rank $\leq 3$ But I am not getting how it can be exactly of rank $2$..