I have some questions about solvable systems. First of all: I'm not a student I just try to understand how this work, so this won't be 100% mathematical correct sorry.
Given is a matrix $\mathbf{A} \in \mathbb{R}^{m \times n}, \mathbf{x} \in \mathbb{R}^{n}$ and $\mathbf{b} \in \mathbb{R}^{m}, \mathbf{b} \neq \mathbf{0}$. ´ Which statements apply to the choice of $A$ and $b$, for which $Ax = b$ has exactly one solution? (original Question of a textbook)
$\mathbf{A}^{T} \mathbf{A} \mathbf{x}=\mathbf{0} \quad \Rightarrow \quad \mathbf{A} \mathbf{x}=\mathbf{0}$
$\|\mathbf{A} \mathbf{x}\|=\left\|\mathbf{A}\left(\mathbf{A}^{\top} \mathbf{A}\right)^{-1} \mathbf{A}^{\top} \mathbf{b}\right\|$
$\|\mathbf{A} \mathbf{x}\|=\left\|\mathbf{A}^{\top} \mathbf{A} \mathbf{x}\right\|$
I want to start with 3. because this is maybe the easy question.
So be $\mathbf{A} \in \mathbb{R}^{m \times n}, \mathbf{x} \in \mathbb{R}^{n}$ with $m = 3$ and $n=2$. Than $Ax$ will be an $2 \times 2$ matrix and $A^TAx$ will be an $1 \times 1$ vector. So the equation of thos two norms can not be correct.
Now to statement 1.:
We know that $b \neq 0$ so $Ax=0$ should be wrong, or am I missing something?
Statement 2. is a problem because I'm not sure how I should operate with norms including matrices.
But maybe I could write it like this
$$\|\mathbf{A} \mathbf{x}\|=\left\|\mathbf{A}\left(\mathbf{A}^{\top} \mathbf{A}\right)^{-1} \mathbf{A}^{\top} \mathbf{b}\right\| = \left\| A((A^T)^{-1} A^{-1})A^Tb \right\|$$
mmh bad Idea, this do not help.
I hope you can help me with those maybe trivial questions :)