I must prove that W1 is a subspace of $R^4$. I am hoping that someone can confirm what I have done so far or lead me in the right direction.
$ W_1 = {(a_1,a_2,a_3,a_4) \in R^4 | 2a_1 – a_2 – 3a_3 = 0 }$
From what I understand, I must show that:
i) The zero vector of $R^4$ is in $W_1$
ii) Sum of any 2 vectors in $W_1$ must also be in $W_1$
iii) any scalar multiple of any vector in $W_1$ must also be in $W_1$
to show i), I just plugged in:
$ 2(0) – (0) – 3(0) = 0 $ therefore we have shown the zero vector is in $W_1$.
for ii), This is what I tried, and I'm not sure if it's correct:
Let $w_1$ and $w_2$ $\in W_1$.
Then I must show that $w_1+w_2 \in W_1$. What I wrote is that since we know by definition of $W_1$ that $w_1 = 0$ and $w_2 = 0$, then $0+0 \in W_1$. Does that prove that $W_1$ is closed under vector addition? If it is I will do something similar for iii), just want to see if I'm on the right track.
Thanks
Best Answer
To show ii:
write $w_1 = (a_1,a_2,a_3,a_4)$ and $w_2 = (b_1,b_2,b_3,b_4)$. We have $$ w_1 + w_2 = (a_1+b_1,a_2+b_2,a_3+b_3,a_4+b_4) $$ Now, you need to check whether, given $w_1,w_2 \in W_1$, we have $$ 2(a_1 + b_1) - (a_2 + b_2) - 3(a_3 + b_3) = 0 $$ Note that we can rewrite the above expression as $$ 2(a_1 + b_1) - (a_2 + b_2) - 3(a_3 + b_3) = (2a_1 - a_2 - 3a_3) + (2b_1 - b_2 - 3b_3) $$