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.

## Guy

19 Oct 2020For 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}.

## Will

2 Aug 2020For #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?

## Jonathan Sharir-Smith

3 Aug 2020This 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!

## Jonathan Sharir-Smith

3 Aug 2020Never mind, I see that this question was asked by Tom C and subsequently answered by Linearity in the affirmative. Will, you are correct.

## Marie

26 Jun 2020Can 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) ?

## Marie

26 Jun 2020*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?

## Linearity

27 Jun 2020Because $6z_1=z_2$ 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).

## Marie

27 Jun 2020this makes so much sense, thanks a lot!

## Jonathan Sharir-Smith

15 May 2020Correct 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".

## Subhasish Mukherjee

20 May 2020Any single vector is vacuously linearly independent. Hopefully that helps?

## Daniel Benvenutti

20 May 2020The point is that the basis of $V=\{0\}$ is not 0, it is ∅. Which is linearly independent and spans $V=\{0\}$.

## Tianze

11 Jun 2020hit the hull's eye

## Tom C

26 Apr 2020I 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?

## Linearity

27 Apr 2020Yes, you are right.

## Shubhadip Ghosh

7 Mar 2020Can (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?

## Linearity

19 Mar 2020Yes, of course.

## Faris

13 Oct 2019In 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).

## Xue Zicheng

5 Mar 2020But you cannot form span｛(1,0,0,0,0),(0,1,0,0,0 )｝in that way.

## samrao1997

21 Apr 2016In 5 I think the last vector of the basis should be x^3 not x^2 because that has degree 2....

## Jiren Jin

8 Jul 2016+1

## 汪哲

8 Dec 2016+1

## Mohammad Rashidi

15 Jan 2017Thank you!