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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.

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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 $\lambda$.

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  • $\begingroup$ Amazing, thank you ! In general that would be the right way to find a basis, right ? Decompose as much as possible as long the the vectors belong to the set. $\endgroup$
    – Kreol
    Commented Aug 3, 2019 at 4:17
  • $\begingroup$ @Kreol Sure. See here for an example given some vectors. In your example, if we allow say, $\pmatrix{1 \\0\\ 0},\pmatrix{0 \\ 1\\ 0},\pmatrix{0 \\ 0\\ 1}$ to be in $S$, then $S$ would no longer be closed under vector addition and scalar multiplication. In general you can find some set of vectors that are in $S$ that span $S$, then check if they are linearly independent by reducing the matrix with rows as said vectors. $\endgroup$
    – Hendrix
    Commented Aug 3, 2019 at 14:18

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