I have been practicing some basic linear algebra and came across this excercise:
Compute the kernel of the mapping $f\colon \mathbb R^{2}\rightarrow \mathbb R, x \mapsto (2,-1)'x$, and draw a picture.
I know what a kernel is and (for most cases) how to compute it, but I have issues understanding the function itself. Is the $(2,-1)'$ a transposed vector (in other exercises the sign for transpose was $x^{t}$)? Should be the function then $f(x)=2x_{1}-x_{2}$? I honestly can not wrap my head around the notation in this problem.
It seems to me that you've got everything right.
Using the symbol $'$ is quite common for transposing a vector (perhaps due to Matlab). And of course the function is just a scalar product between $(2,-1)$ and $x$ $$ f(x)=\left\langle \left[\begin{smallmatrix}2\\-1\end{smallmatrix}\right],x\right\rangle=2x_1-x_2 $$
I assume you know how to proceed from here.