Anticommutators of the spin-1 representation

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The discussion focuses on the anticommutators of the spin-1 representation using the Pauli matrices defined as ##T_{1}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}##, ##T_{2}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix}##, and ##T_{3}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}##. The calculated anticommutators are ##\left\{T_{1},T_{2}\right\}=i\begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}##, ##\left\{T_{1},T_{3}\right\}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}##, and ##\left\{T_{2},T_{3}\right\}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & i \\ 0 & -i & 0 \end{pmatrix}##. The user seeks to derive a general relation from these specific anticommutator results, indicating a deeper exploration of the algebraic structure involved.

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I need to find a closed form of the anticommutators of the Pauli matrices in the spin-1 representation.
The Pauli matrices of the spin-1 representation are given by: ##T_{1}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}##, ##T_{2}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix}## and ##T_{3}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}##. I need to find what ##\left\{T_{i},T_{j}\right\}## is equal to.
Doing some calculations, I found that ##\left\{T_{1},T_{2}\right\}=i\begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}##, ##\left\{T_{1},T_{3}\right\}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}##,##\left\{T_{2},T_{3}\right\}=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & i \\ 0 & -i & 0 \end{pmatrix}##.
Is there a general relation that I can derive from these special relations? I think that I am close, but I can't quite see it.
 
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What else than the explicit matrices do you expect to get? They are as closed from as you can make it.
 

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