infinitesimal generator of SU(3)

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

• Do your $X_7$ and $X_8$ span the same subspace as the ones you find on the Internet? Mar 7, 2020 at 9:22
So, just a minor change of basis, $$X_1=\lambda_1 \qquad X_2=\lambda_4 \qquad X_3=\lambda_6 \qquad X_4=\lambda_2 \\ X_5=\lambda_5 \qquad X_6=\lambda_7 \qquad X_7=-\lambda_3 \qquad X_8=-\frac{1}{2}(\lambda_3 +\sqrt{3} ~ \lambda_8).$$