Feedback to Basis and Dimension of $A:= \{x\in\mathbb{R}^4 | x_2 + 3x_3= 0, x_1=x_2\}$

Question

My question: Because of $\mathbb{R}^4$ I assume the vector has to have $x_1,x_2,x_3,x_4$ To calculate the basis I assume that $x_1= 1$ because $x_1=x_2 as$ the problem says and $x_4= 0$ because this was not given. I got this vector $v= (1,1,3,0)$. Are my assumptions wrong? Should $x_1=0$ because it was not given? My question is about the value of x1. Is it correct that the value of x1 ist 1 because x1 = x2? I just would like to know if I did something wrong. Basis and dimension I can do on my own. Thank you

Answer 1

There are some mistakes, so let me point those out first before giving a solution:

1. I don't understand why you would assume that $x_1 = 1$, nothing here would suggest that.
2. "$x_4 = 0$ because it was not given". No, if there are no $x_4$'s in your equations, you can set $x_4$ to whatever you want.

Here's how you should approach it.If we can isolate the variables, it'll be easier to find out what the equations describe. $x_1=x_2$ is fine allready. We can rewrite $x_2 + 3x_3 = 0$ to $x_2 = -3x_3$. From this, we can rewrite your equations as

$x_2=x_1$ and $x_3=-\frac{1}{3}x_2 = -\frac{1}{3}x_1$

So, when you pick some value for $x_1$, the value of $x_2$ and $x_3$ are allready given. Therefore, we get

$A = \{(a,a,-\frac{1}{3}a, b) | a,b \in \mathbb{R}\}$