Consider the real vector space $\mathbb R^3$ and define $\Psi:\mathbb R^3\rightarrow\mathbb R^3$ by


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 $x$?

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


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


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