Let $x^*\in\text{argmin }f(x)=\text{argmin }\frac{1}{2}\|Ax-b\|^2$ where $A$ is a linear operator. Show that $\text{argmin }f=x^*+\text{Null}(A)$. For $x\in x^*+\text{Null}(A)$ we have $x\in\text{argmin }f$. How to show converse, that for any $x\in\text{argmin }f$, $x$ can be decomposed as $x=x^*+x_0,x_0\in\text{Null}(A)$?
P.S. I accidentally deleted my previous page.
Basically, just plug everything in. Assume we have that
$$ \bar{x} \in \arg\min_x\{ \frac{1}{2}\|Ax-b\|^2\} $$
and we have $\tilde{x} \in Null(A)$. Then,
$$ \frac{1}{2}\|A(\bar{x}+\tilde{x})-b\|=\frac{1}{2}\|A\bar{x} + A\tilde{x} -b\| = \frac{1}{2} \|A\bar{x}-b\|. $$
Here, we used the linearity of $A$ and the assumption that $\tilde{x}\in Null(A)$. Since $\bar{x}+\tilde{x}$ yields the same objective value as $\bar{x}$, we must have that $\bar{x}+\tilde{x} \in \arg\min_x\{ \frac{1}{2}\|Ax-b\|^2\}$ for any $\tilde{x}\in Null(A)$.