$X=(I+ab^T)A(I+ba^T)$;
$A$ 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 $2$..
Best Answer
We can write $X - A$ as $$ X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T $$ Noting that $\operatorname{rank}(PQ) \leq \min\{\operatorname{rank}(P),\operatorname{rank}(Q)\}$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.
It now remains to be shown that the rank is not $1$ or $0$.
First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product $$ b^T A[I + ba^T] $$ is not the zero matrix. To that end, we note that $$ (b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b \neq 0 $$ From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.