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## 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.

### This Post Has 22 Comments

1. For a more formal proof of #1:
Problem: Find all vector spaces that have only one base.
Sol: Let V=U+W be a vector space that is sum of two bases. Also suppose that these two bases are actually one, so U=W. If you extract U from the first equation, U=-W. Now solve this system of equations:
U=W and U=-W, you get U=W={0}.

2. For #8, can one not simply use the fact that each u and each w has a unique representation as a linear combination of the u_m's and the w_n's, and each v has a unique representation as a sum of a u and a w...so that each v therefore has a unique representation as a sum of u_ms and w_ns?

1. This is how I did it also. Each step is justified by 2.29 in the book. I am curious to hear what Linearity has to say about this approach, but no matter what thanks so much for making such an awesome blog!

1. Never mind, I see that this question was asked by Tom C and subsequently answered by Linearity in the affirmative. Will, you are correct.

3. Can someone show how we could write any element of U, (z1,z2,z3,z4,z5) as a linear combination of the basis (1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1) ?

1. *Or rather, can somebody explain the approach taken to obtaining this answer for the basis? I kind of just guess and checked but I am wondering if there is a concrete way to solve such questions?

2. Because $6z_1=z_2$ and $z_3+2z_4+3z_5=0$, we can write $(z_1,z_2,z_3,z_4,z_5)$ as
\begin{align*}
&\ z_1(1,6,0,0,0)-z_4(0,0,2,-1,0)-z_5(0,0,3,0,-1)\\=&\ (z_1,6z_1,-2z_4-3z_5,z_4,z_5).
\end{align*}The point is to replace $z_2$ and $z_3$ with $6z_1$ and $-2z_4-3z_5$, respectively. And then put $$z_1=1,\quad z_4=z_5=0;$$ $$z_4=-1,\quad z_1=z_5=0;$$ $$z_5=-1,\quad z_1=z_4=0;$$to get these vectors (1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1).

1. this makes so much sense, thanks a lot!

4. Correct me if I'm wrong but, for question 1, why isn't the answer no vector spaces. The definition of a basis is a list of vectors in V that span V and are linearly independent. But the only vector in {0} is the zero vector which is not linearly independent. So doesn't that then fail too? Admittedly, this largely rests on the (arbitrary) definition of basis, with emphasis on "in V".

1. Any single vector is vacuously linearly independent. Hopefully that helps?

2. The point is that the basis of $V=\{0\}$ is not 0, it is ∅. Which is linearly independent and spans $V=\{0\}$.

1. hit the hull's eye

5. I love your website! It has been very helpful while I'm doing self-study with Axler. Please correct me if I'm wrong (I'm new at this). Isn't there a shorter way to do #8 by using definition 1.40, which implies that since U + W is a direct sum, for every v in V, there is a unique representation v = u + w. But since u1, ..., um is a basis for U we can express u in that previous expression as a unique combination of the basis vectors (likewise for w). Thus we can express any v in V as a unique combination of the u1, ..., um, w1, ..., wn, which implies by 2.29 that they are a basis for V. It's nice to see a proof straight from the basis definition 2.27, but this other way works too right?

1. Yes, you are right.

6. Can (3,1,0,0,0),(0,1,0,0,0),(0,0,7,1,0),(0,0,0,1,0),(0,0,0,0,1) be a basis with respect to the question number 3.b, as well?

1. Yes, of course.

7. In 4b I think the last vector could also be (0,0,0,1,0) or (0,0,0,0,1) (someone correct me if I'm wrong).

1. But you cannot form span｛(1,0,0,0,0),(0,1,0,0,0 )｝in that way.

8. In 5 I think the last vector of the basis should be x^3 not x^2 because that has degree 2....

1. +1

2. +1

3. Thank you!