How to find a basis for the vector space $P = \lbrace(x,y,z) \in \mathbb{C}^3 : 2x + 2y + 2 iz = 0\rbrace$ over $\mathbb C$?
I believe this is what is called a hyperplane, is that correct?
And how can I check that it actually spans $P$?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Yes, it is a hyperplane. And you do it exactly as you would have done it over the reals. For instance, since$$P=\{(-y-iz,y,z)\mid y,z\in\mathbb C\},$$then $P=\bigl\langle\{(-1,1,0),(-i,0,1)\}\bigr\rangle$, and, since the set $\{(-1,1,0),(-i,0,1)\}$ is linearly independent, it is a basis of $P$.
answered Mar 28, 2020 at 10:34
-
$\begingroup$ Thank you! May I ask how you show that the basis spans the hyperplane? $\endgroup$ Mar 28, 2020 at 12:58
-
$\begingroup$ Because$$(-y-iz,y,z)=y(-1,0,1)+z(-i,0,1).$$ $\endgroup$ Mar 28, 2020 at 13:47
-