1
$\begingroup$

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 $x$?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.