Let $F=\left\{(x, y, z) \in \mathbb{R}^{3}: x+y+z=0\right\}$ be a subspace of $\mathbb{R}^{3}$. Determine a basis of $F$ and then the dimension of $F$.
. We write: $F=\left\{(x, y, z) \in \mathbb{R}^{3}: x=-y-z, y=y, z=z\right\}=$ $\{(-y-z, y, z), y, z \in \mathbb{R}\}=\{(-y, y, 0)+(-z, 0, z), y, z \in \mathbb{R}\}=\operatorname{Vect}\{(-1,1,0),(-1,0,1)\}$. This writing allows us to deduce that the family $\{(-1,1,0),(-1,0,1)\}$ is a generating set of $F$. We also notice that this family is linearly independent. Thus, it is a basis of $F$. Since it contains 2 elements, we conclude that $F$ is of dimension 2 (it is then a vector plane).