# Help finding the Kernel of $\Psi (x_1,x_2,x_3)=(x_1+x_3,0,x_2+x_3)$

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 $$x$$?

• I've edited your question as you have just defined the classical real vector space $\mathbb R^3$ with normal, component-wise, addition. – blub Apr 10 at 21:11
• Thank you @blub – Albert Diaz Apr 10 at 21:19

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 $$\mathrm{ker}(\Psi)=\{(-c,-c,c)\mid c\in\mathbb R\}$$.