I'm going through a least squares computation where $A=\begin{bmatrix}3&2\\5&3\\4&5\end{bmatrix}$ and $\vec{b}=\begin{bmatrix}5\\9\\2\end{bmatrix}$.
From $\vec{x^*}=(A^TA)^{-1}A^T\vec{b}$, I end up getting that $|| \vec{b}-A\vec{x}^*||=0$, but I cannot see how $\vec{b}$ would be in the image of A.
Is it a scalar multiple of one of the column vectors? I've tried multiplying and dividing by various fractional scalars, but to no avail.
First, $x=(A'A)^{-1}A'b$ does not in general imply $|b-Ax|=0$. You can check directly that $$ b-Ax=b-A(A'A)^{-1}A'b=[I-A(A'A)^{-1}A']b $$ where the expression inside the square brackets can be seen as the orthogonal projection onto the space orthogonal to the column space of $A$. Thus, $b-Ax=0$ iff $b$ belongs to the column space of $A$. Another way to see this is by directly using the non-negativity of the Euclidean norm: $$ |b-Ax|=0\iff b-Ax=0\iff Ax=b $$ iff, again, $b$ belongs to the column space of $A$.
In your particular case, $b$ is indeed in the column space of $A$ as you can verify directly that $Ax=b$ where $x=(3,-2)'$. Let's emphasize again that in general, $x=(A'A)^{-1}A'b$ does not entail $|b-Ax|=0$.