Show that $T_{ijk}V_k$ is a second rank tensor in 3D space

Question

The task is show that $$T_{ijk}V_k$$ is a rank-2 tensor.

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Using $$T_{ijk} = a_{i\alpha}a_{j\beta}a_{k\gamma}T_{\alpha\beta\gamma}' $$ and $$V_{k} = a_{k\gamma}V_{\gamma}' $$

I arrive at

$$T_{ijk}V_k = a_{i\alpha}a_{j\beta}a_{k\gamma}T_{\alpha\beta\gamma}'a_{k\gamma}V_{\gamma}'$$

I think the key is to show that $$(a_{k\gamma})^2 = \delta^{k}_{\gamma}$$

but I can't think of how it would work. To me it seems that $$(a_{k\gamma})^2 = 3 $$ which I get from the orthonormality of the rows/columns (with other rows/columns) of the rotation matrices (I'm working in 3D cartesian space) and the double sum over $k$ and $\gamma$, which then gives a (seemingly out of place) factor of 3.

If someone could explain what I'm missing or the correct method to show this, that would be greatly appreciated. Thanks.

Answer 1

You get only one $\gamma$. For the transformation of the second $k$ you need another, independent index. So $$ T_{ijk}V_k = a_{i\alpha}a_{j\beta}a_{k\gamma}T_{\alpha\beta\gamma}'a_{k\eta}V_{\eta}'. $$ Now use that the metric used is euclidean and that the transformation matrix is orthogonal to get $$ a_{k\gamma}a_{k\eta}=\delta_{\gamma\eta}. $$

As a result, only the actually free indices get transformed.