Chapter 2 Exercise B


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.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *