If the linear system $Ax=0$ ($A_{3\times 3}(\mathbb{R})$) has infinitely many solutions, does it mean that the system $Ax=b$ is consistent $\forall b\in\mathbb{R^3}$. If not, find the set of $b\in\mathbb{R^3}$ for which the system $Ax=b$ is consistent.
What characterization of Kronecker Capelli's theorem can prove or disprove this statement?
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Consider $A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$. Then, $x = \begin{bmatrix}0\\0\\t\end{bmatrix}$ is a solution to $Ax = 0$ for any $t \in \mathbb{R}$.
Thus, $Ax = 0$ has infinitely many solutions.
However, since $Ax = \begin{bmatrix}x_1\\x_2\\0\end{bmatrix}$, there is no $x \in \mathbb{R}^3$ such that $Ax = \begin{bmatrix}0\\0\\1\end{bmatrix}$.
Therefore, $Ax = b$ is not consistent for $b = \begin{bmatrix}0\\0\\1\end{bmatrix}$.
For $Ax = b$ to have a solution, we need $b$ to be in the span of the columns of $A$.
This is false. I leave it to you to explicitly find a counterexample. Let me instead describe why you should believe it is false.
Note that $Ax = 0$ having infinitely many solutions in particular implies that the nullspace $N$ has dimension larger than 0. By the Dimension Theorem, this implies that the rank of $A$, which is the dimension of the range, is strictly smaller than 3. Intuitively then, $A$ takes $\mathbf{R}^3$ into $\mathbf{R}^3$ but not to all of $\mathbf{R}^3$ because its dimension is smaller! It will be either a line, plane, or zero. In any case, you can find vectors not reached by $A$.
What can we conclude then? Well, if $b$ is such that $\exists x$ $Ax = b$, then we can say there are infinitely many solutions.
Take $x_H \in N$ be any solution to the homogeneous system $Ax_H = 0$. Then
$$ A(x+x_H) = Ax + Ax_H = Ax = b$$