What is the dimension of a subspace containing $(1,1,1)$?

Question

Say $S=\{{(x,y,z):x=y=z}\}$, then every vector in $s \in S$ can be represented as $\lambda \cdot (1,1,1)$ with $\lambda \in R$. Vector $(1,1,1)$ is trivially independent since it is not zero. Also, $sp(\{(1,1,1)\})=S$. This tells us that $(1,1,1)$ is a basis of $S$ and $\dim (S)=1$. However, I can continue the decomposition: $$ s = \begin{pmatrix}\lambda\\0\\0\end{pmatrix} + \begin{pmatrix}0\\\lambda\\0\end{pmatrix}+\begin{pmatrix}0\\0\\\lambda\end{pmatrix}$$ These vectors are obviously independent and span $S$ as well. So $\dim (S) = 3$, but we just saw that $\dim (S) = 1$. What is the right answer ? Thank you.

Answer 1

The vectors $\begin{pmatrix}\lambda\\ 0\\ 0\end{pmatrix}$, $\begin{pmatrix}0\\ \lambda\\ 0\end{pmatrix}$, $\begin{pmatrix}0\\ 0\\ \lambda\end{pmatrix}$ are not in $S$ for nonzero $\lambda$.