Consider the real vector space $\mathbb R^3$ and define $\Psi:\mathbb R^3\rightarrow\mathbb R^3$ by
$$\Psi(x_1,x_2,x_3)=(x_1+x_3,0,x_2+x_3)$$
How do I got about calculating what the kernel of $\Psi$ is? I am not quite sure I understand what is required.
I am also asked to find 4 elements in $\mathbb R^3$ that are in $\mathrm{ker}(\Psi)$
My attempt: The definition of $\mathrm{ker}(\Psi)=\{g\in\mathbb R^3\mid\Psi(g)=(0,0,0)\}$, since $(0,0,0)$ is the identity in the range $\mathbb R^3$.
Does this mean $\mathrm{ker}(\Psi)=\{(-x,-x,x)\mid x\in\Bbb{R}\}$? Since, by definition $\Psi(-x,-x,x)=(-x+x,0,-x+x)=(0,0,0)$?
As for examples I could just choose any real number for $x$?
As you rightly stated $$\mathrm{ker}(\Psi)=\{x\in\mathbb R^3\mid\Psi(x)=(0,0,0)\}$$
Thus, to characterize $\mathrm{ker}(\Psi)$, we want to verify which $x=(x_1,x_2,x_3)\in\mathbb R^3$ satisfy $\Psi(x)=(0,0,0)$. By the definition of $\Psi$, $\Psi(x)=(0,0,0)$ is equivalent with
$$x_1+x_3=0\\x_2+x_3=0$$
as the second component of $\Psi(x)$ is $0$ anyway.
Solving this system of linear equations, we get that $\Psi(x)=(0,0,0)$ if and only if $x=(-c,-c,c)$ for some $c\in\mathbb R$.
So, concluding, we have $\mathrm{ker}(\Psi)=\{(-c,-c,c)\mid c\in\mathbb R\}$.