[proofplan] We reduce to the [Normal Equations](/theorems/501) and show that the matrix $A^\top A$ is invertible when $A$ has full column rank. The key observation is that full column rank makes $A^\top A$ positive-definite (not merely semi-definite), which guarantees invertibility and yields the explicit formula $x^* = (A^\top A)^{-1}A^\topb$. [/proofplan] [step:Reduce to the normal equations and show $A^\top A$ is positive-definite] By the [Normal Equations](/theorems/501), $x^*$ minimises $\|Ax - b\|^2$ if and only if $A^\top A\,x^* = A^\topb$. It remains to show that $A^\top A$ is invertible when $\operatorname{rank}(A) = n$. For any $x \in \mathbb{R}^n$, $x^\top A^\top Ax = \|Ax\|^2$. If $\operatorname{rank}(A) = n$, then the columns of $A$ are linearly independent, so $Ax = \mathbf{0}$ implies $x = \mathbf{0}$. Therefore $x^\top A^\top Ax > 0$ for all $x \neq \mathbf{0}$, which means $A^\top A$ is positive-definite and hence invertible. [/step] [step:Write down the unique solution] Since $A^\top A$ is invertible, the normal equations $A^\top A\,x^* = A^\topb$ have the unique solution \begin{align*} x^* = (A^\top A)^{-1}A^\topb. \end{align*} [/step]