infinitesimal generator of SU(3)

Question

I have computed generator infinitesimal of SU(3), i got 8 generator $$ X_1= {\begin{pmatrix} 0& 1&0\\ 1& 0&0\\0&0&0 \end{pmatrix}}\\ X_2= {\begin{pmatrix} 0& 0&1\\ 0& 0&0\\1&0&0 \end{pmatrix}}\\ X_3= {\begin{pmatrix} 0& 0&0\\ 0& 0&1\\0&1&0 \end{pmatrix}}\\ X_4= {\begin{pmatrix} 0& -i&0\\ i& 0&0\\0&0&0 \end{pmatrix}}\\ X_5= {\begin{pmatrix} 0& 0&-i\\ 0& 0&0\\i&0&0 \end{pmatrix}}\\ X_6= {\begin{pmatrix} 0& 0&0\\ 0& 0&-i\\0&i&0 \end{pmatrix}}\\ X_7= {\begin{pmatrix} -1& 0&0\\ 0& 1&0\\0&0&0 \end{pmatrix}}\\ X_8= {\begin{pmatrix} -1& 0&0\\ 0& 0&0\\0&0&1 \end{pmatrix}}\\ $$ but apparently when i check in the internet, the last 2 generator , that is $X_7$ and $X_8$ is wrong i calculate it by defining tranformation $T$ near the identity and so on. my question is why the last two generator is wrong

Answer 1

Your matrices are just fine, and just as linearly independent as the eight Gell-Mann matrices.

So, just a minor change of basis, $$ X_1=\lambda_1 \qquad X_2=\lambda_4 \qquad X_3=\lambda_6 \qquad X_4=\lambda_2 \\ X_5=\lambda_5 \qquad X_6=\lambda_7 \qquad X_7=-\lambda_3 \qquad X_8=-\frac{1}{2}(\lambda_3 +\sqrt{3} ~ \lambda_8). $$