1. Find all vector spaces that have exactly one basis.

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. Verify all the assertions in Example 2.28.

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. (a) Let $U$ be the subspace of $\mathbb R^5$ defined by \[U=\{(x_1,x_2,x_3,x_4,x_5)\in \mathbb R^5:x_1=3x_2\text{~and~}x_3=7x_4\}.\]Find a basis of $U$. (b) Extend the basis in part (a) to a basis of $\mathbb R^5$.(c) Find a subspace $W$ of $\mathbb R^5$ such that $\mathbb R^5=U\oplus W$.

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. (a) Let $U$ be the subspace of $\mathbb C^5$ defined by \[U=\{(z_1,z_2,z_3,z_4,z_5)\in \mathbb C^5:6z_1=z_2\text{~and~}z_3+2z_4+3z_5=0\}.\]Find a basis of $U$. (b) Extend the basis in part (a) to a basis of $\mathbb C^5$. (c) Find a subspace $W$ of $\mathbb C^5$ such that $\mathbb C^5=U\oplus W$.

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. Prove or disprove: there exists a basis $p_0$, $p_1$, $p_2$, $p_3$ of $\mathcal{P}_3(F)$ such that none of the polynomials $p_0$, $p_1$, $p_2$, $p_3$ has degree 2.

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. Suppose $v_1,v_2,v_3,v_4$ is a basis of $V$. Prove that \[v_1+v_2,v_2+v_3,v_3+v_4,v_4\] is also a basis of $V$.

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. Prove or give a counterexample: If $v_1,v_2,v_3,v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1,v_2\in U$ and $v_3\notin U$ and $v_4\notin U$, then $v_1,v_2$ is a basis of $U$.

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. Suppose $U$ and $W$ are subspaces of $V$ such that $V=U\oplus W$. Suppose also that $u_1,\cdots,u_m$ is a basis of $U$ and $w_1,\cdots,w_n$ is a basis of $W$. Prove that \[u_1,\cdots,u_m,w_1,\cdots,w_n\] is a basis of $V$.

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.