# Why $SU(3)$ has eight generators?

The generators of $SU(3)$ group are Gell-Mann matrices and one can construct these generators from Pauli spin matrices, basically expanding in 3d and rotating about each axis. Take $\sigma_3$, assume I rotate it about $y$-axis in 3d $$\lambda_{3}=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$ However there is no such Gell-Mann matrix. Instead of it, Gell-Mann $\lambda_8$ $$\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$$ appears which look like a combination of more than one states? I'd appreciate it if anyone explain why don't we have $\sigma_3$ for the rotation about $y$- and $z$-axis, but instead $\lambda_8$? That's why we have finally have 8 eight generators instead 9. I suppose the answer will enlighten the number of generators as well.

• The $\lambda_3$ you have can be obtained by linear combinations of the other generators. Commented May 11, 2015 at 18:17
• Another hint: How many real dimensions does the vector space $su(3)$ of traceless Hermitian $3\times 3$ matrices have? Commented May 11, 2015 at 18:23
• @Qmechanic $n^2-1$ dimension. But still I don't get the $\lambda_8$ and a way of evaluating it. Commented May 11, 2015 at 18:42

Any linear transformation wrought on the Lie algebra of a Lie group yields a valid Lie algebra as I think you understand (the Gell Mann matrices are actually $i$ times the skew-symmetric Lie algebra members), and your proposed $\lambda_3$ is a linear combination of the Gell Mann matrices. The basis comprising $i$ times the Gell Mann matrices does indeed span $\mathfrak{su}(3)$: there are eight of 'em, they are linearly independent and the algebra of $3\times3$ skew-symmetric matrices has eight parameters so they have to be valid!
All group theoretic study could in theory be done with any such Lie algebra. You could use your proposed matrix instead of $\lambda_8$ and theoretically Lie theory will work perfectly. So your question is actually a question about convention: why do we choose this particular basis?
One convenience afforded by the Gell Mann matrices that would not be afforded by your proposed scheme is that the Gell Mann matrices are orthogonal with respect to the Killing form and they are also orthogonal with respect to the ordinary trace inner product (this is because $SU(3)$ is has no continuous center and its Lie algebra stays the same when mapped by the adjoint representation, so that the trace inner product is the same as the Killing form). There can only be two diagonal matrices in the algebra because we can only choose two out of three diagonal elements at will when the matrices concerned need to be traceless, so, if $\lambda_3$ is in the basis, your proposed matrix cannot be: the second diagonal matrix needs to be orthogonal to $\lambda_3$.