If the dimension of $\textrm{ker}(T_A)$ is 2, does the equation $Ax = [1,2,3]^\textrm{T}$ have infinitely many solutions $x$ in $\mathbb{R}^5$? Let $A$ be a $3$x$5$ matrix over $\mathbb{R}$ and let $T_A$ be the associated linear transformation. If the dimension of $\textrm{ker}(T_A)$ is 2, does the equation
$$Ax = \begin{pmatrix}
1\\2\\3
\end{pmatrix}$$
 have infinitely many solutions $x$ in $\mathbb{R}^5$?
Maybe this is in fact a very simple question, but I want to make sure I understand it. Since the dimension of the kernel is 2, the dimension of the range is 3, and consequently $T_A$ will have three pivot points. Does this mean that there are two "free" variables, and hence infinitely many solutions? Is there a more concrete way to put this? Should I be looking at it in terms of the determinant being equal to zero instead?
Any help appreciated!
 A: By the rank-nullity theorem, the image of the linear transformation has dimension $5-2=3$, so must be the whole of $\mathbb{R}^3$. Hence $[1,2,3]^T = Av$ for some $v \in \mathbb{R}^5$. But since the kernel has dimension $2$, it is (in particular) infinite, so for any $w \in \mathrm{ker}(T_A)$ we must have
$$A(v+w)=Av+Aw=[1,2,3]^T+0=[1,2,3]^T$$
So the equation $Ax=[1,2,3]^T$ does indeed have infinitely many solutions.
If $\mathrm{ker}(T_A)$ had dimension $3$ instead, then the image would be a $2$-dimensional subspace of $\mathbb{R}^3$ (i.e. a plane), and the equation $Ax=[1,2,3]^T$ would either have no solutions or infinitely many solutions, depending on whether or not $[1,2,3]^T$ was in the image of $T_A$.
A: You are on the right track. Since $\ker(A)=2$, the range of $A$ has dimension 3, by what is sometimes called the rank-nullity theorem. Thus, the range of $A$ is all of $\mathbb{R}^3$.
Now to show that there are infinitely many solutions to the equation in question.  Clearly $\ker(A)$ contains infinitely many points.  Given $v\in\ker(A)$, we have
$$A(x+v)=Ax+Av=Ax,$$
so if $x$ is a solution to the equation in question, so is $x+v$.
If you have any questions, please let me know and I will edit.
