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Consider, for example, the set of of all symmetric, traceless $4 \times 4$ matrices. I'm trying to find a correctly normalized basis for this set. So far, I have

$$s(1)=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(2)=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(3)=\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right); $$ $$s(4)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(5)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right);s(6)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right); $$ $$ s(7)=\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);s(8)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);s(9)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right); $$ $$s(10)= \frac{1}{\sqrt{6}}\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \\ \end{array} \right), $$

Is this correct? If yes, these are a basis for the $10$-dimensional representation of $SU(4)$, but how do I need to normalize these matrices propely in the convention, where the Dynkin index $Tr(T_a T_b)$ of the lowest-dimensional representation is $\frac{1}{2} \delta_{ab}$?

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    $\begingroup$ The space of symmetric $4 \times $ matrices is $10$-dimensional; the space of traceless symmetric $4 \times 4$ matrices is $9$-dimensional. Note that $$s(10) = -\tfrac{1}{\sqrt{6}}\bigl(s(7) + s(8) + s(9)\bigr).$$That aside, each matrix satisfies $\operatorname{Tr}(T_{a}T_{a}) = 2$, not $\frac{1}{2}$, so it appears each needs to be multiplied by $\frac{1}{2}$. Beyond that, I'm afraid I can't help.... $\endgroup$ – Andrew D. Hwang Jan 3 '16 at 15:09
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    $\begingroup$ @AndrewD.Hwang Ah of course! It turns out this was my main problem and you solved it. Thanks! $\endgroup$ – jak Jan 3 '16 at 15:13
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    $\begingroup$ According to Wikipedia, you want traceless Hermitian $4 \times 4$ matrices (a.k.a., the Lie algebra of $SU(4)$). The space of these is $15$-dimensional: The six complex entries above the diagonal are "free", and three of the four real diagonal entries are free. The corresponding basis matrices are $s(1)$, ..., $s(6)$; $i$ times corresponding skew-symmetric matrices; and $s(7)$, $s(8)$, $s(9)$. Again, $s(10)$ is redundant. $\endgroup$ – Andrew D. Hwang Jan 4 '16 at 16:06

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