Show that $F = \{(a,b,c,d) \in \mathbb{R}^4 : a-2b = 0 \land b+c = 0\}$ is a subspace of $\mathbb{R}^4$

Question

$F = \{(a,b,c,d) \in \mathbb{R}^4 : a-2b = 0 \land b+c = 0\}$

In a previous example in my book what is done is that the given values (a,b,c,d) are rewritten so that all have the same variable. You show that it's a linear combination by multiplying a vector by that variable and because all the linear combinations of vectors of a subspace are part of that subspace, that proves this... for example: $$F = \{(a,b,c) \in \mathbb{R}^4 : a+2c = 0 \land b+c = 0\} \Leftrightarrow\\ \{(-2c,-c,c):c\in\mathbb{R}\} \\ =\{c(-2,-1,1):c \in \mathbb{R}\}$$

I tried to do the same thing in this case and got as far as: $$F = \{(-2c,-c,c,\frac{c}{c}d):c,d \in\mathbb{R}\} \\ =\{c(-2,-1,1,\frac{1}{c}d):c,d\in \mathbb{R}\}$$

Is this enough proof or is more necessary?

Answer 1

Your solution is not correct, since you have divided by $c$, but $c$ might well be $0$.

Note that if $(a,b,c,d),(a',b',c',d')\in F$, then$$(a,b,c,d)+(a',b',c',d')=(a+a',b+b',c+c',d+d'),$$which belongs to $F$, since$$(a+a')-2(b+b')=a-2b+a'-2b'=0+0=0$$and$$b+b'+c+c'=b+c+b'+c'=0+0=0.$$On the other hand, if $\lambda\in\mathbb R$, then$$\lambda(a,b,c,d)=(\lambda a,\lambda b,\lambda c,\lambda d),$$which also belongs to $F$, by a similar argument.

So, since clearly $F\neq\emptyset$, $F$ is a subspace of $\mathbb R^4$.

Answer 2

You only need the following two facts:

• for any vector $u\in\Bbb R^n$, $S_u=\{x:x.u=0\}$ is a subspace of $\Bbb R^n$;
• the intersection of two subspaces is a subspace.

Now just take $u_1=(1,-2,0,0)$, and $u_2=(0,1,1,0)$, so that $$F=S_{u_1}\cap S_{u_2}$$