Vector sets subspaces

Question

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 :)

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

1. The set of all linear combinations of the vectors $(1,0,1,0)$ and $(0,1,0,1)$ (of $\mathbb R^4$).
2. The set of all vectors of the form $(a,b,a -b,a+b)$ (of $\mathbb R^4$).
3. The set of all vectors $(x,y,z)$ such that $x+y+z=0$ (of $\mathbb R ^3$).

Answer 1

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 $\mathbb R^3.$

Answer 2

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.

Answer 3

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.

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(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.

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(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.

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(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.