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$$\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})\psi$$

and

$$A^{a}_{\mu} \to A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c}$$

induce the gauge transformation

$$F_{\mu\nu}^{a} \to F_{\mu\nu}^{a} - f^{abc}\theta^{b}F_{\mu\nu}^{c}.$$

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Proof:

Let us begin with the definition of the field-strength tensor in terms of the covariant derivative ##D_{\mu} = \partial_{\mu}-iA_{\mu}^{a}T^{a}_{\bf R}##:

$$F^{a}_{\mu\nu}T^{a}_{\bf R} = i\left[D_{\mu},D_{\nu}\right].$$

In order to determine the gauge transformation of ##F^{a}_{\mu\nu}T^{a}_{\bf R}##, we must first determine the gauge transformation of ##D_{\mu}D_{\nu}\psi##. It is rather easy to show that

$$D_{\mu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}\psi$$

and hence

$$D_{\mu}D_{\nu}\psi \to (1 \pm i\theta^{a}T^{a}_{\bf R})D_{\mu}D_{\nu}\psi.$$

Therefore, we now have

$$F_{\mu\nu}^{a}T^{a}_{\bf R} = i\left[D_{\mu},D_{\nu}\right]$$

$$F_{\mu\nu}^{a}T^{a}_{\bf R} \to i(1 \pm i\theta^{b}T^{b}_{\bf R})\left[D_{\mu},D_{\nu}\right]$$

$$F_{\mu\nu}^{a}T^{a}_{\bf R} \to (1 \pm i\theta^{b}T^{b}_{\bf R})F_{\mu\nu}^{a}T^{a}_{\bf R}$$

$$F_{\mu\nu}^{a}T^{a}_{\bf R} \to F_{\mu\nu}^{a}T^{a}_{\bf R} \pm i\theta^{b}T^{b}_{\bf R}F_{\mu\nu}^{a}T^{a}_{\bf R}$$

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In the last line of the derivation, I don't have a third term needed to form a structure constant. Where have I gone wrong?