Prove that the lie algebra $\mathfrak{so}(3)$ does not admit any two-dimensional Lie subalgebra.
Consider the basis given by $e_1=\pmatrix{0&1&0\\-1&0&0\\0&0&0}, e_2=\pmatrix{0&0&1\\0&0&0\\-1&0&0}$ and $e_3=\pmatrix{0&0&0\\0&0&1\\0&-1&0}$ for $\mathfrak{so}(3)$.
Suppose that $\mathfrak g$ is a Lie subalgebra of $\mathfrak{so}(3)$, with basis $$\{v=v^1e_1+v^2e_2+v^3e_3, u=u^1e_1+u^2e_2+u^3e_3\}.$$
Now I think that the idea is to show that the Lie bracket is not closed i.e. $[v,u] \notin \mathfrak g$?
Computing I got that $$[v,u]=(v^2u^3-v^3u^2)e_1+(v^3u^1-v^1u^3)e_2+(v^1u^2-v^2u^1)e_3$$ but I'm not sure why this isn't in $\mathfrak g$?
Hint, but deliberately not a complete answer:
Suppose that it were in $g$. Then it would be a linear combination of the first two (nonzero) elements, right? So what would be the determinant of $$ \pmatrix{v^1 & v^2 & v^3 \\ u^1 & u^2 & u^3 \\ v^2u^3-v^3u^2&v^3u^1-v^1u^3&v^1u^2-v^2u^1}? $$
And what would that tell you about the coefficient vectors $u$ and $v$?