Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to prove/disprove that $W$ is a linear subspace, and I'm not sure my approach is correct (especially the last point I'm making). Please correct me if I'm wrong.

Let $V$ be a set of vectors over $F=\mathbb{R}$, $V=\mathbb{R}^4$ and $W$ is a subgroup of $V$ such that $$W=\{(x,y,z,w)\in V|(x+y)^2=0\}$$

My solution:

For $x=0 , y=0$ we'll get $(x+y)^2=(0+0)^2=0$, thus $0\in W$.

Checking if $W$ is close under vector addition: $\forall w_1 , w_2 \in W. w_1=(x_1.-x_1,0,0) , w_2=(x_2,-x_2,0,0)|\big(x_1+x_2+(-x_1)+(-x_2)\big)^2=(0)^2=0\in W$

Checking if $W$ is close under scalar multiplication: Given $W\subseteq \mathbb{R}^4 ,\alpha\in \mathbb{R}$ we can conclude that $\forall w\in W|\alpha \times w \subseteq \mathbb{R}^4$, therefore $\forall a\in F, w\in W|\alpha \times w\in W$.

Therefore $W$ is a linear subspace of $V$.

share|cite|improve this question
Might it be helpful also to note that $(x+y)^2=0$ is equivalent to $x+y=0$? – Daryl Apr 16 '13 at 21:23
up vote 5 down vote accepted

You should put $v_1 = (x_1, y_1, z_1, w_1) \in W$ and $v_2 = (x_2, y_2, z_2, w_2)\in W$ and determine whether $v_1 + v_2 = (x_1+x_2, y_1+y_2, z_1 + z_2, w_1+ w_2)\in W$.

Of course, only the $x_i, y_i$ matter in determining whether the sum is in $W$, but you want to show that the sum of any two vectors in $W$ is again in $W$, not just those with $z_i, w_i = 0$. The mere existence of vectors in $W$ whose sum is again in $W$ isn't the point. To prove closure, we need to show that for any (read every) two vectors in W, their sum is again in $W$. If there exist vectors in W whose sum is NOT in $W$, then closure fails.


Is it always the case that for $v_1 + v_2 = (x_1+x_2, y_1+y_2, z_1 + z_2, w_1+ w_2)\in W$, $$(x_1+x_2)^2 + (y_1 + y_2)^2 = x_1^2 + 2x_1x_2 + x_2^2 + y_1^2 + 2y_1y_2 + y_2^2 = 0?$$ If and only if so, then do you have closure. We know that $x_1^2 + y_1^2 = 0,$ and that $x_2^2 + y_2^2 = 0$, but can you see why there may be problems with closure?

BTW: I did a double take on your proof of closure: ending with "$ 0 \in W$" makes it look like the zero vector is in W, which you showed immediately above.

share|cite|improve this answer
Thanks, Amzoti. It was a slow day for me today...feeling a tad ill. :-( – amWhy Apr 17 '13 at 1:02
me too, and I have a 1.5 talk tomorrow! Been a very slow day on MSE for me today, although I just gave a crypto answer and that is also a favorite topic! Feel better and make sure to rest! Regards – Amzoti Apr 17 '13 at 1:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.