Exercise 5.4.5 (Minimal norm solution)

Question

$
ewcommand{\T}{^\mathsf{T}} 
ewcommand{\mtbf}[2]{\mathbf{#1}_{#2}}$ Let an equation $A\mathbf{x = b}$ has a solution, and let $A$ has non-trivial kernel (so the solution is not unique). Prove that 
\begin{enumerate}[label = \alph*)]
    \item There exists a unique solution $\mtbf{x}{0}$ of $A\mathbf{x = b}$ minimizing the norm $||\mathbf{x}||$, i.e., that there exists unique $\mtbf{x}{0}$ such that $A \mtbf{x}{0} = \mathbf{b}$ and $||\mtbf{x}{0}|| \leqslant ||\mathbf{x}||$ for any $\mathbf{x}$ satisfying $A\mathbf{x = b}$.
    \item $\mtbf{x}{0} = P_{(	ext{Ker} \, A)^\perp} \mathbf{x}$ for any $\mathbf{x}$ satisfying $A \mathbf{x=0}$. 
\end{enumerate}

Jing Huang · Accepted Answer

a) Suppose $\mtbf{x}{0}, \mtbf{x}{1}$ are solutions of $A \mathbf{x = b}$. Then $A(\mtbf{x}{1} - \mtbf{x}{0}) = \mathbf{0}$. i.e., $\mtbf{x}{1} - \mtbf{x}{0} \in 	ext{Ker} \, A$. As a result, $P_{(	ext{Ker} \, A)^\perp} (\mtbf{x}{1} - \mtbf{x}{0}) = \mathbf{0} = P_{(	ext{Ker} \, A)^\perp} \mtbf{x}{1} - P_{(	ext{Ker} \, A)^\perp} \mtbf{x}{0}$. So we have $P_{(	ext{Ker} \, A)^\perp} \mtbf{x}{1} = P_{(	ext{Ker} \, A)^\perp} \mtbf{x}{0} = const := \mathbf{h}$.  \par
Note that $||\mathbf{x}||^2 = ||P_{(	ext{Ker} \, A)^\perp} \mathbf{x}||^2 + ||\mathbf{x} - P_{(	ext{Ker} \, A)^\perp} \mathbf{x}||^2 \geqslant ||\mathbf{h}||^2$ for any $\mathbf{x}$ satisfying $A \mathbf{x = b}$. When $\mtbf{x}{0} - P_{(	ext{Ker} \, A)^\perp} \mtbf{x}{0} = \mathbf{0}$, $\mtbf{x}{0} = \mathbf{h}$, such a $\mtbf{x}{0}$ has the smallest norm among all the solutions. The existence and uniqueness of $\mtbf{x}{0}$ are guaranteed by $\mathbf{h}$.  \par
b) It is shown above.

Exercise 5.4.5 (Minimal norm solution)

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Exercise 5.4.5 (Minimal norm solution)

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