Proving that a set is a basis in two different ways During a lecture, my teacher solved the following exercise:

Let $B$ be a set of linearly independent vectors in vector space $V$: $B = \{ v_1, v_2, v_3 \}$.
Prove that $B' = \{ v_1+v_2, v_2-v_3, v_1+v_2+v_3 \}$ is a basis of $\operatorname{Span}(B)$.

The teacher solved this by using coordinate vectors – he took the vectors from $B'$ as coordinate vectors using the $B$ basis, put them in a matrix (as columns) like so:
$$\begin{bmatrix}1&0&1\\1&1&1\\0&-1&1\end{bmatrix}$$
and by showing that this matrix is invertible, meaning the vectors are linearly independent, and knowing that $B$ has 3 vectors in it, therefore $\dim\operatorname{Span}(B)=3$, you prove that this is indeed a basis for $B$.
However, isn't it much simpler to just say something like:

The vectors in $B'$ are linear combinations of the vectors in $B$, meaning their span stays the same (since elementary operations don't change the span), and of course there are 3 vectors and we know that $\dim\operatorname{Span}(B)=3$. Isn't this a quicker, simpler approach? Or am I missing something?

Thanks!
 A: Let $B''=\{v_1+v_2,v_1-v_3,v_2+v_3\}$. The second vector in $B''$ is a linear combination of the first and third, so $\dim\operatorname{span}B''=2$ and $B''$ is not a basis for $\operatorname{span}B$.
By your argument, $B''$ would be a basis for $\operatorname{span}B$, but it is not.
A: It's great that you're trying to find another proof to this problem. We just have to clear up a few things. You stated

The vectors in B' are linear combinations of the vectors in B, meaning their span stays the same (since elementary operations don't change the span)

Let's start with the parenthetical statement. I assume you're talking about elementary column/row matrix operations: 1) swapping columns/row, 2) multiplying columns/rows by a constant 3) add a multiple of one column/row to another column/row. If we view the matrix as a set of column/row vectors then yes, as you said, these operations (so long as the constant is nonzero in operation 2) do not change the span of these column/row vectors. [Be careful, if they are row vectors you must do row operations and similarly for column vectors you must do column operations].
If you see the operations your professor has done as taking the original matrix $B$ (read as a column matrix let's say) and getting $B'$  by adding column 2 to column 1, subtracting column 3 from column 2, and adding columns 1 and 2 to column 3, then yes, based on this fact that elementary matrix operations (excluding zeroing out a vector) does not change the span, your proof would be correct. 
However, stating that the vectors in $B'$ are linear combinations of vectors in $B$ is a much weaker statement and this is what the other folks in this thread are talking about. In the extreme I could say $B'$ is $\{0v_1, 0v_2, 0v_3\}$ and it is definitionally a linear (albeit trivial) combination of vectors in $B$ but we all know that its span is the zero vector. The other have provided other examples of where saying a linear combination gives the same span is false. 
Bottom line, you have good intuition and made a correct connection with elementary operations, but the jump to a statement about all linear combinations does not work.
