Is $\{(a, b, c) \in \mathbb{C}^3 : a^3 = b^3\}$ a subspace of $\mathbb{C}^3$ I have two questions to solve:


*

*Is $\{(a, b, c) \in \mathbb{R}^3 : a^3 = b^3\}$ a subspace of $\mathbb{R}^3$?

*Is $\{(a, b, c) \in \mathbb{C}^3 : a^3 = b^3\}$ a subspace of $\mathbb{C}^3$?


For the first one, I proved it is. Then for the second part, I found almost no difference. So I am not sure if I am on the right track. I mean, I believe there must be some differences. But I can't figure it out.
 A: As somebody else commented, the difference in $\mathbb{C}$ is that $a^3 = b^3$ does not imply $a = b$. Essentially, complex numbers have multiple different cube roots, just like real numbers have two different square roots (e.g. $\sqrt{1}$ can be $-1$ or $1$).
To disprove that the set in question is a subspace of $\mathbb{C}$, we can provide a counterexample of two complex vectors that are members of the set, while their sum is not.
Let $x = (e^{i\frac{π}{3}}, -1, 0)$ and $ y = (1,1,0)$.
Its sum won't satisfy the property $a^{3}=b^{3}$
(Note: $e^{i\frac{π}{3}}$ is a shorthand for $cos(\frac{π}{3}) + sin(\frac{π}{3})i$, which is a point on the unit circle drawn in the complex plane, see this wonderful video for an in-depth explanation of complex numbers and how they relate to e, this should make it intuitively clear why taking the cube of that number is $-1$.)
$(e^{i\frac{π}{3}})^3 = e^{3*i\frac{π}{3}} = e^{iπ} = -1 = (-1)^3$, so x is in the set, and y is obviously in the set, too.
But $x+y = (1+e^{i\frac{π}{3}}, 0, 0)$ is not in the set, because $(1+e^{i\frac{π}{3}})^3 \neq 0$. Therefore, the set is not a subspace of $\mathbb{C}^3 $because it is not closed under addition.
A: As somebody else commented, the difference in $\mathbb{C}$ is that $a^3 = b^3$ does not imply $a = b$. Essentially, complex numbers have multiple different cube roots, just like real numbers have two different square roots (e.g. $\sqrt{1}$ can be $-1$ or $1$).
To disprove that the set in question is a subspace of $\mathbb{C}$, we can provide a counterexample of two complex vectors that are members of the set, while their sum is not:
Let $x = (e^{i\frac{π}{3}}, -1, 0)$ and $ y = (1,1,0)$.
(Note: $e^{i\frac{π}{3}}$ is a shorthand for $cos(\frac{π}{3}) + sin(\frac{π}{3})i$, which is a point on the unit circle drawn in the complex plane, see this wonderful video for an in-depth explanation of complex numbers and how they relate to e, this should make it intuitively clear why taking the cube of that number is $-1$.)
$(e^{i\frac{π}{3}})^3 = e^{3*i\frac{π}{3}} = e^{iπ} = -1 = (-1)^3$, so x is in the set, and y is obviously in the set, too.
But $x+y = (1+e^{i\frac{π}{3}}, 0, 0)$ is not in the set, because $(1+e^{i\frac{π}{3}})^3 \neq 0$. Therefore, the set is not a subspace of $\mathbb{C}^3 $because it is not closed under addition.
