Evaluation of the spin-sum for a massive spin-1 particle?

Question

Assuming that a massive spin-1 particle has momentum only in the z-direction, the polarization vectors are given by

$$\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$

$$\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0, \frac{E}{m})$$

$$\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$

The so-called spin-sum is the claimed to be

$$
\sum\limits_{J_z = -1,0,+1} \varepsilon_{\mu}\varepsilon_{\nu}^* = g_{\mu\nu} + \frac{p_{\mu}p_{\nu}}{m^2}
$$

I absolutely dont understand how this spin-sum is evaluated.
What does $\varepsilon_{\mu}\varepsilon_{\nu}^*$ even exactly mean? Is it a scalar product between two of the three above polarization vectors, or is it a "tensor-product" between the components of a single polarization vector which results in a 4x4 matrix and one has finally to sum all such matrices for the three possible values of $J_z$?
I would really appreciate it if somebody can explain to me what this spin-sum exactly means and how it is evaluated step-by-step.869

Comments

It is a tensor-product, as you could have seen from the right hand side. Just evaluate the three matrices explicitly and sum them to get the result. It is pure linear algebra.

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To add to Arnold, here might be a even straighter way for OP to see this is a elementary linear algebra calculation: Think of $\varepsilon_\mu(J_z)$ as a 4 by 3 matrix $\varepsilon$, rows labeled by $\mu$ and columns labeled by $J_z$, then the spin sum is nothing but $\varepsilon \varepsilon^\dagger$, which is a 4 by 4 matrix.