Find the basis and dimension of $M=\{(x_1,x_2,x_3,x_4) \in \Bbb R^{4}: x_1-x_3-2x_4=0, x_1+x_2-2x_3=2x_4\}$.
I think I would know how to do it myself, however I don't understand what exactly am I doing and why. First I write
$(x_1, x_2, x_3, x_4)=(x_3+2x_4, x_3,x_3, x_4)$
$(x_1, x_2, x_3, x_4)=x_3(1,1,1,0)+x_4(2,0,0,1)$.
So $(1,1,1,0)$ and $(2,0,0,1)$ form the basis. The dimension is $2$.
I don't know if this is correct, and please can someone explain in simple terms, what is the idea behind all this? Could I have chosen $2$ vectors that are in $M$, check that they are linearly independent and say they form the basis?
From the equations defining $M$, we have that $$ x_1=x_3+2x_4,\quad x_2=-x_1+2x_3+2x_4=-(x_3+2x_4)+2x_3+2x_4=x_3.$$ Hence any vector in $M\subset \mathbb{R}^4$ can be represented as $$(x_3+2x_4,x_3,x_3,x_4)=x_3(1,1,1,0)+x_4(2,0,0,1).$$ Since for any $\alpha\in\mathbb{R}$, $(2,0,0,1)\not=\alpha (1,1,1,0)$, $(1,1,1,0)$, and $(2,0,0,1)$ are linearly independent and, as you said, they are a basis for $M$.
The answer is yes also for the other question, but be careful because $(2,5,3)$ and $(1,3,2)$ are not in $M$ (they are in $\mathbb{R}^3$!). Note that $$u:=a(1,1,1,0)+b(2,0,0,1),\quad v:=c(1,1,1,0)+d(2,0,0,1)$$ is a base for $M$ as soon as $ad\not=bc$ (why?). So another base of $M$ is: $(3,1,1,1)$, $(4,2,2,1)$.