1. Solution: The only vector spaces is $\{0\}$. For if there is a nonzero vector $v$ in a basis, then we can get a new basis by changing $v$ to $2v$.

Here, we just consider the fields $\mathbb R$ and $\mathbb C$, hence $v\ne 2v$. For finite fields such as $\mathbb{F}_2$, there may be other solutions.

2. We have already known how to check linear independence. So, we just verify they are a spanning list. The process is easy and tedious, so I omit them.

3. Solution:

(a) $(3,1,0,0,0)$, $(0,0,7,1,0)$ and $(0,0,0,0,1)$.

(b) $(3,1,0,0,0)$, $(0,0,7,1,0)$, $(0,0,0,0,1)$, $(1,0,0,0,0)$ and $(0,0,1,0,0)$.

(c) $W=\mathrm{span}\{(1,0,0,0,0),(0,0,1,0,0)\}$ by (b).

4. Solution:

(a) $(1,6,0,0,0)$, $(0,0,2,-1,0)$ and $(0,0,3,0,-1)$.

(b) $(1,6,0,0,0)$, $(0,0,2,-1,0)$, $(0,0,3,0,-1)$, $(1,0,0,0,0)$ and $(0,0,1,0,0)$.

(c) $W=\mathrm{span}\{(1,0,0,0,0),(0,0,1,0,0)\}$ by (b).

Here I consider the vector space is over $\mathbb C$.

5. Solution: Because $1$, $x$, $x^2$, $x^3$ is a basis of $\mathcal{P}_3(F)$, hence \[ 1+x^3,x+x^3,x^2+x^3,x^3 \]is also a basis of $\mathcal{P}_3(F)$. However none of the polynomials $1+x^3$, $x+x^3$, $x^2+x^3$, $x^3$ has degree 2.

Here I use a fact that $v_1,v_2,v_3,v_4$ is a basis of $V$, then\[v_1+v_4,v_2+v_4,v_3+v_4,v_4\] is also a basis of $V$. Proof of this is similar to Problem 6.

6. Solution: First, we need show that $v_1+v_2,v_2+v_3,v_3+v_4,v_4$ is linear independent. Assume that \[0=a(v_1+v_2)+b(v_2+v_3)+c(v_3+v_4)+dv_4,\] then $av_1+(a+b)v_2+(b+c)v_3+(c+d)v_4=0$. Note that $v_1,v_2,v_3,v_4$ is a basis of $V$, it follows that $a=0$, $a+b=0$, $b+c=0$ and $c+d=0$. Then $a=b=c=d=0$, this means $v_1+v_2,v_2+v_3,v_3+v_4,v_4$ is linear independent.

Now, note that \[v_3=(v_3+v_4)-v_4,\quad v_2=(v_2+v_3)-(v_3+v_4)+v_4\]and \[v_1=(v_1+v_2)-(v_2+v_3)+(v_3+v_4)-v_4,\]we can conclude that $v_1,v_2,v_3,v_4$ can be expressed as linear combinations of $v_1+v_2,v_2+v_3,v_3+v_4,v_4$. Hence all vectors that can be expressed as linear combinations of $v_1,v_2,v_3,v_4$ can also be linearly expressed by $v_1+v_2,v_2+v_3,v_3+v_4,v_4$, i.e. $v_1+v_2,v_2+v_3,v_3+v_4,v_4$ spans $V$.

Above all, \[v_1+v_2,v_2+v_3,v_3+v_4,v_4\] is also a basis of $V$.

7. Solution: Counterexample: let $V=\mathbb R^4$ , $v_1=(1,0,0,0)$, $v_2=(0,1,0,0)$, $v_3=(0,0,1,1)$, $v_4=(0,0,0,1)$ and \[U=\{(x,y,z,0)|x,y,z\in\mathbb R\}.\]Then all the conditions are satisfied, but $v_1,v_2$ is not a basis of $U$ since $(0,0,1,0)$ can not be linearly expressed by $v_1,v_2$.

8. Solution: First, we show that $u_1,\cdots,u_m,w_1,\cdots,w_n$ is linearly independent. If there exist $a_1,\cdots,a_m\in\mathbb F$ and $b_1,\cdots,b_n\in\mathbb F$ such that \[a_1u_1+\cdots+a_mu_m+b_1w_1+\cdots+b_nw_n=0.\]Then \[a_1u_1+\cdots+a_mu_m=-(b_1w_1+\cdots+b_nw_n)\in U\cap W,\]it follows that \[a_1u_1+\cdots+a_mu_m=0,\quad b_1w_1+\cdots+b_nw_n=0\]since $V=U\oplus W$ implies $U\cap W=\{0\}$. However, note that $u_1,\cdots,u_m$ is a basis of $U$ and $w_1,\cdots,w_n$ is a basis of $W$, it follows that $a_1=\cdots=a_m=0$ and $b_1=\cdots=b_n=0$. Hence $u_1,\cdots,u_m,w_1,\cdots,w_n$ is linearly independent.

Now, it suffices to verify that $u_1,\cdots,u_m,w_1,\cdots,w_n$ spans $V$. For any $v\in V$, there exist $u\in U$ and $w\in W$ such that $v=u+w$ since $V=U\oplus W$. Note that $u_1,\cdots,u_m$ is a basis of $U$ and $w_1,\cdots,w_n$ is a basis of $W$, it follows that there exist $a_1,\cdots,a_m\in\mathbb F$ and $b_1,\cdots,b_n\in\mathbb F$ such that \[u=a_1u_1+\cdots+a_mu_m,\]\[w=b_1w_1+\cdots+b_nw_n.\]Hence \[v=u+w=a_1u_1+\cdots+a_mu_m+b_1w_1+\cdots+b_nw_n,\]which means $u_1,\cdots,u_m,w_1,\cdots,w_n$ spans $V$.

Above all, \[u_1,\cdots,u_m,w_1,\cdots,w_n\] is a basis of $V$.

Remark: For any $v\in U\cap W$, if I can show that $v=0$. Then $U\cap W=0$, since we choose $v$ arbitrarily. I have always skipped this step. For instance, for any $v\in V$, if I can show that $v\in W$, then $V\subset W$. The argument is similar.