Rank of Matrix Determination $X=(I+ab^T)A(I+ba^T)$;
$A$ 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 ?
Efforts:
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$
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 $2$..

$(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form 
\begin{equation}
 P = Z(Z^TZ)^{-1}Z^T
\end{equation}
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$, 
\begin{equation}
 Z = 
 \begin{bmatrix}
  a & Ab
 \end{bmatrix}
\end{equation}
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that 
\begin{equation}
 Pa = a \tag{1}
\end{equation}
and 
\begin{equation}
 PAb = Ab \tag{2}
\end{equation}
Hence 
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is 
$$P(X-A) =  Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) =  ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that 
\begin{equation}
 (I-P).(X-A) = 0
\end{equation}
where $I-P$ is of rank $n-2$.
