Prove set is a basis of a vector space Let $\{x,y\} = \{(x_{1}, x_{2}), (y_{1}, y_{2})\}$ be a basis of $K^{2}$. Prove that for a scalar $a \in K$, $a\neq 0$, the set $\{ x+y, ax\} = \{(x_{1}+y_{1}, x_{2}+y_{2}), (ax_{1}, ax_{2})\}$ is also a basis of $K^{2}$.
I'm trying to prove this using the fact that since both sets have the same dimension, I can prove that the second set is linearly independent and therefore a basis of $K$.
Since $\{x,y\}$ is a basis of $K^{2}$ then there's $\alpha_{1}, \alpha_{2} \in K$ so that $\alpha_{1}x_{1}+\alpha_{2}y_{1}=0$ and $\alpha_{1}x_{2}+\alpha_{2}y_{2}=0$, and to prove that $\{x+y, ax\}$ is linearly independent I have to show that there's $\beta_{1}, \beta_{2} \in K$ so that $(\beta_{1}+\beta_{2}a)x_{1} + \beta_{1}y_{1}=0$ and $(\beta_{1}+\beta_{2}a)x_{2} + \beta_{1}y_{2}=0$.
Is it valid if I say that $\alpha_{1}=\beta_{1}+\beta_{2}a$ and $\alpha_{2}=\beta_{1}$?
 A: This is rather confused...
Yes, since you have a set with two elements in a $2$-dimensional space, it is enough to show the set is linearly independent in order to show it is a basis. But what you write after that is incorrect.
You say that because the elements $(x_1,x_2)$ and $(y_1,y_2)$ form a basis of $K$ (you mean $K^2$, I think), then there exist $\alpha_1,\alpha_2\in K$ such that $\alpha_1 x_1+ \alpha_2 y_1=0$ and $\alpha_1 x_2+\alpha_2 y_2=0$... that would be such that $\alpha_1(x_1,x_2)+\alpha_2(y_1,y_2)=(0,0)$... Well, such scalars always exist whether what you have is a basis or not... namely, $\alpha_1=\alpha_2=0$. The fact that your set is a basis implies, in fact, that these are the only values of $\alpha_1,\alpha_2$ that work.
Then what you say you need to prove to show that $\{x+y,ax\}$ is linearly independent is likewise incorrect. You do not need to prove the existence of scalars as you describe (again, $\beta_1=\beta_2=0$ will work but won't tell you anything about linear independence). Rather, you need to show that the only solution to
$$\beta_1(x_1+y_1,x_2+y_2) + \beta_2(x_1,x_2) =(0,0)$$
is $\beta_1=\beta_2=0$. Yes, you will need to use that $x$ and $y$ are linearly independent, but you need to know what that means, because it is not what you twice state.
Rather it looks like you might be trying to prove the set spans from knowing the original spans. That's valid, but your description is again incorrect. You would need to show that for every $r,s\in K$ there exist $\beta_1,\beta_2\in K$ such that
$$\beta_1(x+y) + \beta_2(ax) = (r,s).$$
You know that there exist $\alpha_1,\alpha_2\in K$ such that $\alpha_1 x + \alpha_2 y =(r,s)$, and you can try to use these to determine the $\beta_i$. This is also valid, but not what you wrote.
I also note that the problem mentions a $b$ that is never used...
A: Seems like you got the definition wrong. To prove that a set is a basis you need to prove two things:

*

*The set is linearly independent

*The set spans the entirety of the vector space

For 1, you could proceed by contradiction: Suppose that $\{x+y, ax\}$ is linearly dependent (i.e. there exist $\alpha_1, \alpha_2$, at least one of them not equal to $0$, such that $\alpha_1(x+y)+\alpha_2x=0$), then rearrange into the form $\beta_1x+\beta_2y=0$. If you can show that at least one of $\beta_1, \beta_2$ is nonzero, this contradicts the premise that $\{x, y\}$ is linearly independent.
For 2, you can proceed more directly: As we know $\{x, y\}$ forms a basis, for any $c\in K^2$, we have $\alpha_1, \alpha_2$ such that $\alpha_1x+\alpha_2y=c$. You can then rearrange to find $\beta_1, \beta_2$ such that $\beta_1(x+y)+\beta_2ax=c$. Since these coefficients exist for any $c$, we can conclude $\{x+y, ax\}$ spans $K^2$.
Once you prove both statements, you will have shown $\{x+y, ax\}$ is a basis.
