Consider the inner product $g_{\mu\sigma} g^{\nu\sigma}$. We see that $\mu$~and $\nu$ select two rows in the determinant; we have to take each element in turn from the $\mu$~row, multiply by the minor of the corresponding element of the $\nu$~row, add together, and divide the result by~$g$. This is equivalent to substituting the $\mu$~row for the $\nu$~row and dividing the resulting determinant by~$g$. If $\mu$~is not the same as~$\nu$ this gives a determinant with two rows identical, and the result is~$0$. If $\mu$~is the same as~$\nu$ we reproduce the determinant~$g$ divided by itself, and the result is~$1$. We write \[ \left. \begin{aligned} g_{\mu}^{\nu} &= g_{\mu\sigma} g^{\nu\sigma} \\ &= 0 \quad\text{if $\mu \neq \nu$} \\ &= 1 \quad\text{if $\mu = \nu$} \end{aligned} \right\}. \Tag{(25.1)} \] Thus $g_{\mu}^{\nu}$~has the same property of a substitution-operator that we found \index{Substitution-operator}% for $\dfrac{\dd x_{\mu}}{\dd x_{\alpha}'}\, \dfrac{\dd x_{\alpha}'}{\dd x_{\nu}}$ in~\Eq{(22.4)}. For example\footnotemark,\footnotetext {Note that $g_{\mu}^{\nu}$~will act as a substitution-operator on \emph{any} expression and is not restricted to operating on tensors.} \[ g_{\mu}^{\nu} A^{\mu} = A^{\nu} + 0 + 0 + 0. \Tag{(25.2)} \] Note that $g_{\nu}^{\mu}$~has not the same meaning as~$g_{\mu}^{\nu}$ with $\mu = \nu$, because a summation is implied. Evidently \[ g_{\nu}^{\nu} = 1 + 1 + 1 +1 = 4. \Tag{(25.3)} \] The equation~\Eq{(25.2)} shows that $g_{\mu}^{\nu}$~multiplied by any contravariant vector always gives a vector. Hence $g_{\mu}^{\nu}$~is a tensor. It is a very exceptional tensor since its components are the same in all coordinate-systems. \PageSep{56} Again since $g_{\mu\sigma}g^{\nu\sigma}$~is a tensor we can infer that $g^{\nu\sigma}$~is a tensor. This is proved rigorously by remarking that $g_{\mu\sigma} A^{\mu}$~is a covariant vector, arbitrary on account of the free choice of~$A^{\mu}$. Multiplying this vector by~$g^{\nu\sigma}$ we have \[ g_{\mu\sigma} g^{\nu\sigma} A^{\mu} = g_{\mu}^{\nu} A^{\mu} = A^{\nu}, \] so that the product is always a vector. Hence the rigorous quotient theorem applies. The tensor character of~$g^{\mu\nu}$ may also be demonstrated by a method which shows more clearly the reason for its definition as the minor of~$g_{\mu\nu}$ divided by~$g$. Since $g_{\mu\nu} A^{\nu}$~is a covariant vector, we can denote it by~$B_{\mu}$. Thus \[ g_{11} A^{1} + g_{12} A^{2} + g_{13} A^{3} + g_{14} A^{4} = B_{1}; \text{ etc.} \] Solving these four linear equations for $A^{1}$, $A^{2}$, $A^{3}$, $A^{4}$ by the usual method of determinants, the result is \[ A^{1} = g^{11} B_{1} + g^{12} B_{2} + g^{13} B_{3} + g^{14} B_{4}; \text{ etc.,} \] so that \[ A^{\mu} = g^{\mu\nu} B_{\nu}. \] Whence by the rigorous quotient theorem $g^{\mu\nu}$~is a tensor.