= Underdetermined minimum-norm example
{title2=$x_*=(3/5,6/5)^T$}
Now take one equation for two unknowns:
$$
A=\begin{pmatrix}1&2\end{pmatrix},
\qquad b=3.
$$
The <linear system> $x_1+2x_2=3$ has an entire <affine line> of exact solutions. Since $A$ has <full row rank>, its <Moore-Penrose inverse> is
$$
A^+=A^T(AA^T)^{-1}
=\frac15\begin{pmatrix}1\\2\end{pmatrix},
\qquad
\boxed{x_*=A^+b=(3/5,6/5)^T.}
$$
The <kernel> is $\operatorname{span}\{(-2,1)^T\}$, and all exact solutions are
$$
x=x_*+t(-2,1)^T,\qquad t\in\mathbb R.
$$
Because $x_*$ is <orthogonal> to $(-2,1)^T$,
$$
\|x\|_2^2=\frac95+5t^2.
$$
The <Euclidean norm> is therefore smallest exactly at $t=0$. The zero residual alone cannot select a solution; the minimum-norm rule supplies the additional criterion.
\Image[/pseudoinverse-minimum-norm-line.png]
{title=The minimum-norm solution is the closest point of the solution line to the origin}
{description=The <affine line> $x_1+2x_2=3$ is perpendicular to $x_*=(3/5,6/5)^T$. Moving along the line changes only a <kernel> component and increases $\|x\|_2^2$ by $5t^2$.}
This selection depends on the specified <inner product>. Rescaling one unknown changes what counts as a small <Euclidean norm>, so units and weighting matter in an application. A minimum-norm answer is mathematically canonical for the chosen geometry, rather than an automatic guarantee of physical plausibility.
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