$A=\frac{xx^T}{x^Tx}$.Find the dimension of the vector space $[y \in \mathbb{R^3}:Ay=0]$

Question

Let $x \in \mathbb{R^3}$ be a non null vector and $A=\frac{xx^T}{x^Tx}$.Find the dimension of the vector space $[y \in \mathbb{R^3}:Ay=0]$. Clearly the question asks what is the dimension of the null space of $A$. We know it is $n-\text{rank(A)}$ So, here we need rank$(A)$. Now, I just made an observation. $x^Tx$ is a scalar. We know $\text{Rank(A)}=\text{Rank}(A^T) = \text{Rank}(x^Tx)=1 $,as it is a scalar.

So, the dimension of the null space of $A=3-1=2$ Am I correct?

Answer 1

You are on the right track. Yet there might be a mistake. You could check that $A^{\top}=A$, especially, $\left(xx^{\top}\right)^{\top}=xx^{\top}$.

Instead, you could make use of the rank inequality, i.e., $$ \text{rank}\left(AB\right)\le\min\left\{\text{rank} A,\text{rank}B\right\}. $$ Therefore, $$ \text{rank}\left(xx^{\top}\right)\le\text{rank}x=1. $$ Thus as long as $xx^{\top}\ne O$, we have $$ \text{rank}\left(xx^{\top}\right)=1. $$ The rest of your proof follows hereafter.

Answer 2

You’re close. The key isn’t that $x^Tx$ is a scalar but that $x^Ty$ is: $Ay = {xx^T\over x^Tx}y = {x^Ty\over x^Tx}x$, so for all $y$, $Ay$ is a scalar multiple of $x$. Since $x\ne0$, this means that the rank of $A$ is one and so the dimension of its null space is $3-1=2$.