Q:

If $\{a,b,c\}$ is a basis for $\Bbb R^3$, is $\{a+b,b-c,c-a\}$ is also a basis for $\Bbb R^3?$

If $\{a,b,c\}$ is a basis for $\Bbb R^3$, is $\{a+b,b-c,c-a\}$ is also a basis for $\Bbb R^3?$

What I thought was to look at the standard basis:
$$a = (1,0,0)$$
$$b = (0,1,0)$$
$$c = (0,0,1)$$
For this basis we get: 
$$a+b = (1,1,0)$$
$$b-c = (0,1,-1)$$
$$c-a = (-1,0,1)$$
To check if the new vectors are linear independent, put them as rows in a matrix and rank, you get: 
$$\begin{bmatrix}1&1&0\\0&1&-1\\0&0&2\end{bmatrix}$$
Namely, the vectors are linear independent. 
Is it true to conclude from this basis, the standard basis, for all the bases in $\Bbb R^3$ regarding the question above? 

A:

Yes it is enough to check that the matrix
$$
C = \begin{bmatrix}1&1&0\\0&1&-1\\-1&0&1\end{bmatrix}
$$ has linearly independent columns/rows. But the reason why this works even though you are just checking a special case is as follows:
To get the new set of vectors $\{a + b, b - c, c - a\}$ from your original basis $\{a, b, c\}$, you can write the basis vectors $a, b, c$ as $1 \times 3$ row vectors and stack them on top of each other to get a $3 \times 3$ matrix:
$$
\begin{bmatrix}a\\b\\c\end{bmatrix}
$$ It looks like a column vector but each row is whole vector so you have a matrix but matrix-matrix multiplication is nice so that you can pretend like it is a column vector when you multiply $C$ to it:
$$
C \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1&1&0\\0&1&-1\\-1&0&1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}a + b\\b - c\\c - a\end{bmatrix}
$$ Well you get your new basis stacked on top of one another as a matrix. Now you know that the matrix $\begin{bmatrix}a\\b\\c\end{bmatrix}$ has linearly independent rows because the rows form a basis. Also you have already checked that $C$ has linearly independent rows. Thus their product $\begin{bmatrix}a + b\\b - c\\c - a\end{bmatrix}$ must also have linearly independent rows i.e. your new set of vectors is a basis.

