Vector sets subspaces Can anyone guide me alone this question? I get the general meaning of subspaces but how do I show that the vectors are subspaces for this question?
Thank you in advance :)

Show that the following sets of vectors are subspaces of $\mathbb R^m$.


*

*The set of all linear combinations of the vectors $(1,0,1,0)$ and $(0,1,0,1)$ (of $\mathbb R^4$).

*The set of all vectors of the form $(a,b,a -b,a+b)$ (of $\mathbb R^4$).

*The set of all vectors $(x,y,z)$ such that $x+y+z=0$ (of $\mathbb R ^3$).

 A: A subset $U\subseteq V$ of $V$ is a subspace iff:


*

*$0\in U$

*for every $u,v\in U$ and every scalar $\alpha$, vector $\alpha u+v$ belongs to $U$ 


Let's check it for the case (c). We take $U=\{(x,y,z)\in \mathbb R^3 : x+y+z=0\}$ and check that:


*

*$0=(0,0,0)\in U$ as $0+0+0=0$

*if $u=(x_1,y_1,z_1)\in U$ and $v=(x_2,y_2,z_2)\in U,$ then $x_1+y_1+z_1=x_2+y_2+z_2=0$,
so 
$$\alpha x_1+x_2+\alpha y_1+y_2+\alpha z_1+z_2=0,$$ 
what means that $\alpha u+v\in U$. Therefore $U$ is a subspace of $\mathbb R^3.$

A: Hint: to check that $W$ is a vector subspace of a real vector space $V$, one should prove that $0 \in W$ and $\lambda w + u \in W$ for any $\lambda \in \mathbb{R}$ and $w,u \in W$. 
So, for example, in the first case: an element of your $W$ is given by $(\mu,\nu,\mu,\nu)$, with $\mu,\nu \in \mathbb{R}$. Then you should prove that 
$$\lambda(\mu,\nu,\mu,\nu)+(\mu',\nu',\mu',\nu') \in W$$
for any $\lambda,\mu',\nu' \in \mathbb{R}$. But 
$$\lambda(\mu,\nu,\mu,\nu)+(\mu',\nu',\mu',\nu') = (\lambda \mu+\mu',\lambda \nu+\nu',\lambda \mu+\mu',\lambda \nu+\nu')$$
and $$(\lambda \mu+\mu',\lambda \nu+\nu',\lambda \mu+\mu',\lambda \nu+\nu') = (\lambda \mu+\mu')(1,0,1,0)+(\lambda \nu + \nu')(0,1,0,1).$$
You can immediately check that $0 \in W$, so the claim is true.
A: I'm calling the sets in question $S_a,S_b,S_c$ respectively.
Since $\mathbf{0}$ is trivially in all of them, they are non-empty.

(a) 
Suppose there exist $u,v\in S_a$ and $a,b,c,d,M,N\in\mathbb{R}$ such that
$$u=a(1,0,1,0)+b(0,1,0,1)\\v=c(1,0,1,0)+d(0,1,0,1)$$
Consider
\begin{align}
Mu+Nv&=M(a(1,0,1,0)+b(0,1,0,1))+N(c(1,0,1,0)+d(0,1,0,1))\\
&=Ma(1,0,1,0)+Nb(0,1,0,1)+Mc(1,0,1,0)+Nd(0,1,0,1)\\
&=M(a+c)(1,0,1,0)+N(b+d)(0,1,0,1)
\end{align}
which is a linear combination of $(1,0,1,0)$ and $(0,1,0,1)$.
Hence, $Mu+Nv\in S_a$ and $S_a$ is closed under vector addition and scalar multiplication.

(b) Suppose there exist $u,v\in S_b$ and $a,b,c,d,M,N\in\mathbb{R}$ such that
$$u=(a,b,a-b,a+b)\\v=(c,d,c-d,c+d)$$
Consider
\begin{align}
Mu+Nv&=M(a,b,a-b,a+b)+N(c,d,c-d,c+d)\\
&=(Ma,Mb,Ma-Mb,Ma+Mb)+(Nc,Nd,Nc-Nd,Nc+Nd)\\
&=(Ma+Nc,Mb+Nd,Ma-Mb+Nc-Nd,Ma+Mb+Nc+Nd)\\
&=(Ma+Nc,Mb+Nd,(Ma+Nc)-(Nd+Mb),(Ma+Nc)+(Mb+Nd))\\
&\in S_b\\
\end{align}
Hence $S_b$ is closed under vector addition and scalar multiplication.

(c) Suppose there exist $u,v\in S_c$ and $x,y,z,p,q,r,M,N\in\mathbb{R}$ such that
$$u=(x,y,z)\\v=(p,q,r)$$
Consider
\begin{align}
Mu+Nv&=M(x,y,z)+N(p,q,r)\\
&=(Mx+Np,My+Np,Mz+Nr)
\end{align}
Note that
$$Mx+Np+My+Np+Mz+Nr=M(x+y+z)+N(p+q+r)=M(0)+N(0)=0$$
Hence, $Mu+Nv\in S_c$ and $S_c$ is closed under vector addition and scalar multiplication.
