Column Space from Least Sqaures

Question

Let's say

$$A = \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 2 & 1 \\ \end{bmatrix}$$

and $A \cdot x = b$

and $ b = \begin{bmatrix} 4\\ -1 \\ -1 \\ \end{bmatrix}$

After calculating $x = (A^TA)^{-1}A^Tb$, we get $x = \begin{bmatrix} 0 \\ 0\\ \end{bmatrix}$

So what does it say about $x$ and $b$?

I think the answer is $b$ is not the column space of $A$ because we cannot find any solution of $X$ to equal to $b$? Is there something to do with projection?

Answer 1

Yes, exactly. $x$ is the solution of the Gauss Normal Equation, which is equivalent to the statements that $x$ is a minimiser of $f(x')=||Ax' - b||$ and $Ax=Pb$, where $P$ is the orthogonal projection from $\mathbb{R}^3$ to $R(A)$ (the image of $A$).