Finding the kernel of a linear transformation mapping from $\Bbb R^2\to\Bbb R$ $g:R^2\to R$ defined via $g(x,y)=x+2y$. Find the kernel of the linear transformation and justify. My thought was to find the coordinates of this space ${[1,0], [0,2]}$ as columns. Then, row reduce to the $2x2$ identity and augment with $[x,y]=[0,0]$. I believe this will give the trivial solution to the transformation, making the kernel $[0,0]$. Is this correct?
 A: You want to find all vectors $(x, y)$ such that $x+2y=0$. Fix the variable $y=1$. Solving for $x$ gives $x=-2$. Thus, the vector $(-2, 1)\in Ker(g)$.
Thus, the kernel of $g$ is at least $1$ dimensional. We know that it can't be two dimensional as that would mean that $Ker(g)=R^2$. And we know that $Ker(g)\ne R^2$ as $g(1, 0)\ne 0$. Thus, we conclude that $Ker(g)$ is one dimensional and the explicit description of $Ker(g)$ is given by $$Ker(g)=\{\lambda (-2,  1) : \lambda \in R\}$$
A: There is a lot of ways to find the kernel of a linear transformation. First, note that if $T \colon V \to W$ is linear, then if $u, v \in \ker T$ and $\alpha, \beta \in \mathbb{R}$, we have $$T(\alpha u + \beta v) = \alpha T(u) + \beta T(v) = 0,$$ so the kernel is always a linear subspace of the domain. From this, a dimensional analysis can solve your problem.
Note that the dimension of the domain is $2$, so the kernel must be of dimensions $2$, $1$ or $0$. It is not of dimension $2$, since this would imply that the kernel is the whole domain, which in fact is only the case if your transformation is null.
It isn't zero also, since $T(2,-1) = 0$ and $(2,-1) \neq (0,0)$. So, it must be of dimension $1$. This means that, to find the kernel, you just have to find a vector that is not zero, but is in the kernel. The kernel will be the subspace generated by this vector.
As I alredy pointed out, $T(2,-1) = 0$, so you just have to take $\ker T = \mathbb{R}(2,-1) = \{(2t, -t) \mid t \in \mathbb{R}\}$.
