$\newcommand{\ones}{\mathbf 1}$

Suppose $A\in {\mathbf R}^{m \times n}$, with $m < n$, is full rank. Let $x_\mathrm{ln} = A^T(AA^T)^{-1}y$. Consider the following statements.

$x_\mathrm{ln}$ satisfies $Ax_\mathrm{ln} = y$.

$x_\mathrm{ln}$ is the unique point in ${\mathbf R}^n$ closest (in terms of norm) to a solution of $Ax=y$.

• Incorrect.
• Correct! There are many solutions of $Ax=y$ that are not least-norm solutions.

$x_\mathrm{ln}$ is the point in $\mathcal N(A)$ that has smallest norm.

If $x$ satisfies $Ax=y$, then $x-x_\mathrm{ln}\in \mathcal N(A)$.

Let $x^\star$ be the mimimizer of $\|Ax-y\|^2 + \mu \|x\|^2$, where $\mu >0$. (Note that $x^\star$ depends on $y$ and $\mu$.)

$x^\star$ is a linear function of $y$.

$x^\star$ is a linear function of $\mu$.