Linear span problems > Determine the dimension of $\operatorname{span}(\underline{a}-\underline{b},\underline{a}+\underline{b}+\underline{c},-2\underline{a}-\underline{c})$, given $\underline{a},\underline{b}$ and $\underline{c}$ are linearly independent. I know that $$\begin{align*}\operatorname{span}( \underline{a}-\underline{b},\underline{a}+\underline{b}+\underline{c},-2\underline{a}-\underline{c}) &= \operatorname{span}( 2\underline{a}+\underline{c},\underline{a}+\underline{b}+\underline{c},-2\underline{a}-\underline{c}) \\\ &= \operatorname{span}( \underline{0},\underline{a}+\underline{b}+\underline{c},-2\underline{a}-\underline{c}) \\\ &=\operatorname{span}( -\underline{a}+\underline{b},-2\underline{a}-\underline{c})\end{align*}$$ But I don't understand why I can conclude that these two are in the basis and therefore the dimension is 2?

Without loss of generality, we may assume that $a,b,c$ is the standard basis. So the vectors in coordinates are given by $$ u=\begin{pmatrix}1 \\\ -1 \\\0 \end{pmatrix}, \;v=\begin{pmatrix}1 \\\ 1 \\\1 \end{pmatrix}, \; w=\begin{pmatrix}-2 \\\ 0 \\\\-1 \end{pmatrix}. $$ We want to determine $n=\dim span(u,v,w)$. Clearly $1\le n\le 3$ by definition here. But because $u+v+w=0$, we have $n\le 2$. However, each two are linearly independent, so $n=2$.