We call the tensor A_{μν} the “extension” of the tensor A_{μ}. Then we can easily show that this combination also leads to a tensor, when the vector A_{μ} is not representable as a gradient. In order to see this we first remark that ψ (_d_φ/∂_x__{μ}) is a co-variant four-vector when ψ and φ are scalars. This is also the case for a sum of four such terms :— $$ S_{\mu} = \psi^{(1)} \frac{\partial \phi^{(1)}}{\partial x_{\mu}} + ... + \psi^{(4)} \frac{\partial \phi^{(4)}}{\partial x_{\mu}} $$ when ψ^{(1)}, φ^{(1)} ... ψ^{(4)}, φ^{(4)} are scalars. Now it is however clear that every co-variant four-vector is representable in the form of S_{μ}. If for example, A_{μ} is a four-vector whose components are any given functions of _x__{ν}, we have, (with reference to the chosen co-ordinate system) only to put ψ^{(1)} = A₁ φ^{(1)} = _x₁_ ψ^{(2)} = A₂ φ^{(2)} = _x₂_ ψ^{(3)} = A₃ φ^{(3)} = _x₃_ ψ^{(4)} = A₄ φ^{(4)} = _x₄_. in order to arrive at the result that S_{μ} is equal to A_{μ}. In order to prove then that A_{μν} is a tensor when on the right side of (26) we substitute any co-variant four-vector for A_{μ} we have only to show that this is true for the four-vector S_{μ}. For this latter case, however, a glance on the right hand side of (26) will show that we have only to bring forth the proof for the case when A_{μ} = ψ ∂φ/∂_x__{μ}. Now the right hand side of (25) multiplied by ψ is $$ \psi \frac{\partial^2 \phi}{\partial x_{\mu} \partial x_{\nu}} - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} \psi \frac{\partial \phi}{\partial x_{\tau}} $$ which has a tensor character. Similarly, (∂φ/∂_x__{μ}) (∂φ/∂_x__{ν}) is also a tensor (outer product of two four-vectors). Through addition follows the tensor character of $$ \frac{\partial}{\partial x_{\nu}} (\psi \frac{\partial \phi}{\partial x_{\mu}}) - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} (\psi \frac{\partial \phi}{\partial x_{\tau}}) $$ Thus we get the desired proof for the four-vector, ψ ∂φ/∂_x__{μ} and hence for any four-vectors A_{μ} as shown above. With the help of the extension of the four-vector, we can easily define “extension” of a co-variant tensor of any rank. This is a generalisation of the extension of the four-vector. We confine ourselves to the case of the extension of the tensors of the 2nd rank for which the law of formation can be clearly seen. As already remarked every co-variant tensor of the 2nd rank can be represented as a sum of the tensors of the type A_{μ} B_{ν}. It would therefore be sufficient to deduce the expression of extension, for one such special tensor. According to (26) we have the expressions $$ \frac{\partial A_{\mu}}{\partial x_{\sigma}} - \begin{Bmatrix}\sigma\mu\\\tau\end{Bmatrix} A_{\tau} $$ $$ \frac{\partial B_{\nu}}{\partial x_{\sigma}} - \begin{Bmatrix}\sigma\mu\\\tau\end{Bmatrix} B_{\tau} $$