Could someone explain or comment the proof X from my lecture notes?
We have been given:
- $A \in \mathbb{C}^{m \times n}$
- $A^+$ a Moore-Penrose-Inverse
- $V$ some unitary space
- $b \in \mathbb{C}^m$
- a system of linear equations $Ax=b$
- the theorem Y that says:
$\varphi: V \to V$ with $\varphi^2=\varphi$ is a self-adjoint operator $\Leftrightarrow \ \forall u \in Im(\varphi) \land \forall w \in Ker(\varphi): \langle u,w \rangle = 0$. In this case we call $\varphi$ an orthogonal projection.
- the theorem Z that says:
if $\varphi$ is an orthogonal projection, then for all $v \in V: \varphi(v)$ is a unique vector from $Im(\varphi)$ that has the minimal distance to $v$.
We want to show:
- for all $x \in \mathbb{C}^n: \Vert Ax-b \Vert \ge \Vert AA^+b-b \Vert$
The proof X:
As $AA^+AA^+=AA^+$ and as $AA^+$ is a hermitian matrix, $AA^+$ defines an orthogonal projection according to the theorem Y. It is $$Im(\varphi) \subseteq \{Ax \mid x \in \mathbb{C}^n\}$$ and for $Ax \in U$: $$Ax=AA^+Ax=\varphi(Ax) \in Im(\varphi).$$ So $\varphi$ is an orthogonal projection on $Im(\varphi)$, and the theorem Z implies the assertion.
- What does "$AA^+$ defines a map" mean?
- What does it mean, "$AA^+$ defines a projection according to Y"? Y says something different, what we do according to it?
- What is $\{ Ax \mid x \in \mathbb{C}^n\}$? If this should be a set of vectors in $\mathbb{C}^m$, why do we write it this way? We have just said that we work with $AA^+$, why do we multiply $x$ with $A$ and construct some set?
- Why $Im(\varphi) \subseteq \{Ax \mid x \in \mathbb{C}^n\}$?
- Why $\varphi(Ax) \in Im(\varphi)$?
- How these two equalities imply that $\varphi$ is an orthogonal projection on $Im(\varphi)$? Was it not that we assumed in the beginning?
- How theorem Z implies the inequality?
