# 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 $2$..

• How to show that X−A is a rank 2 matrix?? – Debasish Jana Sep 20 '18 at 15:05
• Welcome to MSE. It is more likely that you will get responses when you share your efforts. – Ahmad Bazzi Sep 20 '18 at 15:06
• $X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... 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.. – Debasish Jana Sep 20 '18 at 15:12
• Also $+1$ for showing your efforts :) – Ahmad Bazzi Sep 20 '18 at 15:15
• @StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$ – Omnomnomnom Sep 20 '18 at 15:27

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 $2$.

• now that's a really nice argument .. – Ahmad Bazzi Sep 20 '18 at 15:35
• Yours is pretty nifty as well, Ahmad. – Omnomnomnom Sep 20 '18 at 15:40

## $(X-A)$ is spanned by two vectors only

We can solve this using projector matrices, i.e. matrices of the form $$P = Z(Z^TZ)^{-1}Z^T$$ The number of columns of $X$ will determine the rank of the matrix $X -A$. If we stack in $Z$, $$Z = \begin{bmatrix} a & Ab \end{bmatrix}$$ We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that $$Pa = a \tag{1}$$ and $$PAb = Ab \tag{2}$$ 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 $$(I-P).(X-A) = 0$$ where $I-P$ is of rank $n-2$.

• Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$. – Omnomnomnom Sep 20 '18 at 15:47