I am struggling with the following Problem: \begin{align}Y= \lbrace (x^4-y^4,0,0,0) \mid x,y \in \mathbb{R} \rbrace \subset \mathbb{R}^4 \end{align} Question, is the given Set a subspace of $\mathbb{R}^4 ?$
(Answer given by my tutor: Yes)
I thought about it as follows \begin{align}(x^4-y^4,0,0,0)=\underbrace{(x^4,0,0,0)}_{:=u}+\underbrace{(-y^4,0,0,0)}_{:=v} \end{align} I defined two new vectors, because the notation $x,y$ is a bit confusing for me (usually the statements only depend on $x$). Then I could say that $u \in Y$ and $v \in Y$ and clearly, the sum would be in $Y$ too. Intuitive that seems fine to me, but I have the daunting feeling that I am being cyclic in this problem or make things too easy for myself.
However, if I continue on this way I would also need to verify that if $u \in Y$ then $\lambda u $ must be in $Y$ too. However, here I see the following the problem, if $u \in Y$ then when $\lambda$ is applied to this vector, this would scale the $x$-component and would contribute to make it not a subspace anymore.
I used WolframAlpha to visualize my idea:
http://www.wolframalpha.com/input/?i=%28x%5E4-y%5E4%2C0%2C0%29+ (before lambda)
http://www.wolframalpha.com/input/?i=%285x%5E4-y%5E4%2C0%2C0%29+ (after lambda being applied)
This just has to do with the fact that $f\colon\mathbb{R}\rightarrow\left[0,\infty\right)$ with $f\left(a\right)=a^{4}$ is onto. Suppose $u,v\in Y$. Then \begin{align*} u+v & =\left(x_{u}^{4}-y_{u}^{4},0,0,0\right)+\left(x_{v}^{4}-y_{v}^{4},0,0,0\right)\\ & =\left(\underbrace{\left(x_{u}^{4}+x_{v}^{4}\right)}_{r_{1}}-\underbrace{\left(y_{u}^{4}+y_{v}^{4}\right)}_{r_{2}},0,0,0\right) \end{align*} Since $r_{1},r_{2}\in\left[0,\infty\right)$, we can take $x=r_{1}^{1/4}$ and $y=r_{2}^{1/4}$ to get $$ u+v=\left(x^{4}-y^{4},0,0,0\right)\in Y. $$