# Category: Chapter 2

## Chapter 2 Exercise C

1. Solution: Let $u_1,u_2,\cdots,u_n$ be a basis of $U$. Thus $n=\dim U=\dim V$. Hence $u_1,u_2,\cdots,u_n$ is a linearly independent list of vectors in V with length $\dim V$. By 2.39, $u_1,u_2,\cdots,u_n$ is a basis...

## 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...

## Chapter 2 Exercise A

1. Solution: We just need to show that $v_1$, $v_2$, $v_3$, $v_4$ can be expressed as linear combination of $v_1-v_2$, $v_2-v_3$, $v_3-v_4$, $v_4$. Note that $v_1=(v_1-v_2)+(v_2-v_3)+(v_3-v_4)+v_4,$$v_2=(v_2-v_3)+(v_3-v_4)+v_4,$$v_3=(v_3-v_4)+v_4,\quad v_4=v_4,$we will get the conclusion by definition 2.17....