Is It A Basis? Span? 
Let $\{\vec{a},\vec{b},\vec{c}\}$ be a basis for $\mathbb{R}^3$
Prove:
  
  
*
  
*$\{\vec{a}+\vec{b},\vec{b}-\vec{c}, \vec{c}-\vec{a}\}$ Spans  $\mathbb{R}^3$
  
*$\{\vec{a}-\vec{b},2\vec{b}+\vec{c}, \vec{a}+\vec{b}+\vec{c}\}$ is a basis for $\mathbb{R}^3$

I am assuming that all $a,b,c$ vectors as they were marked in underscore in the book.


*

*$\{\vec{a}+\vec{b},\vec{b}-\vec{c}, \vec{c}-\vec{a}\}=\{\vec{a},\vec{b}, \vec{c}\}+\{\vec{b},-\vec{c},-\vec{a}\}$


And $\{\vec{a},\vec{b},\vec{c}\}\subseteq Span(\{\vec{a},\vec{b}, \vec{c}\}+\{\vec{b},\vec{c},-\vec{a}\})$


*We need to prove that the $3$ vectors are linearly independent 
$\{\vec{a}-\vec{b},2\vec{b}+\vec{c}, \vec{a}+\vec{b}+\vec{c}\}=\{\vec{a},2\vec{b}\,\vec{a}\}+\{-\vec{b},\vec{c},\vec{b}\}+\{0,0,\vec{c}\}$  which are linearly independent.
Assuming $a,b,c$ scalars


*

*$\{a+b,b-c,c-a\}=a\{1,0,-1\}+b\{1,1,0\}+c\{0,-1,1\}$ row reducing the matrix 


$\begin{pmatrix} 
1 & 0 & -1 \\
1 & 1 & 0\\
0 & -1 & 1 \\
\end{pmatrix}\sim \begin{pmatrix} 
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1 \\
\end{pmatrix}$ so it spans $\mathbb{R}^3$


*$\{a-b,2b+c,a+b+c\}=a\{1,0,1\}+b\{-1,2,1\}+c\{0,1,1\}$ row reducing the matrix 


$\begin{pmatrix} 
1 & 0 & 1 \\
-1 & 2 & 1\\
0 & 1 & 1 \\
\end{pmatrix}\sim \begin{pmatrix} 
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1 \\
\end{pmatrix}$ 
So we have $3$ vectors that are linearly independent so it is a basis of $\mathbb{R}^3$
Does it make any sense? was a bit confused from this qeustion  
 A: If you're allowed to use the fact that three linearly independent vectors span a 3-dimensional space, then you only need to show that the vectors in the following sets of parts 1. and 2. are indeed linearly independent, given the linear independence of $\{\vec{a},\vec{b},\vec{c}\}$.

  
*
  
*$\{\vec{a}+\vec{b},\vec{b}-\vec{c}, \vec{c}-\vec{a}\}$ Spans  $\mathbb{R}^3$
  
*$\{\vec{a}-\vec{b},2\vec{b}+\vec{c}, \vec{a}+\vec{b}+\vec{c}\}$ is a basis for $\mathbb{R}^3$

To show linear independence of $\{\vec{a}+\vec{b},\vec{b}-\vec{c}, \vec{c}-\vec{a}\}$, write:
$$\alpha ( \vec{a}+\vec{b} ) + \beta ( \vec{b}-\vec{c} ) +\gamma ( \vec{c}-\vec{a} ) = 0$$
Rearranging, this is equivalent to:
$$ ( \alpha - \gamma )\vec{a} +  ( \alpha + \beta )\vec{b} + ( \gamma - \beta )\vec{c} = 0$$
Linear independence of $\{\vec{a},\vec{b},\vec{c}\}$ now gives:
$$\left\{ \begin{array}{rcl}
\alpha - \gamma &=& 0 \\
 \alpha + \beta &=& 0 \\
 \gamma - \beta &=& 0
\end{array}\right.
\implies
\left\{ \begin{array}{rcl}
\alpha  &=& 0 \\
  \beta &=& 0 \\
 \gamma  &=& 0
\end{array}\right.$$
So $\{\vec{a}+\vec{b},\vec{b}-\vec{c}, \vec{c}-\vec{a}\}$ is a set of three linearly independent vectors in $\mathbb{R}^3$ and thus spans $\mathbb{R}^3$.
This is easier than explicitly showing that this set spans $\mathbb{R}^3$, but you can do that as well.
A: Starting from a basis you can construct others basis by linear combination of the vectors of the given basis.
You only have to check that these linear combinations are linearly independent working on the coefficient as you did.
EG in the first you have the following vectors of coefficient for the linear combination 
(1,0,-1), (1,1,0), (0,-1,1)
and these vectors are linearly independent.
Another’s way to see this is to think in therms of matrix for the change of basis. To obtain a basis the matrix of the change must ne not singular.
