The tensor product of nondegenerate real symmetric forms has multiplicative signature

By [F3] there are bases $v_1,\dots ,v_n$ of $V$ and $w_1,\dots ,w_m$ of $W$ diagonalizing $B$ and $C$: $B(v_i,v_{i^{\mathrm{\prime }}})=b_i{\delta }_{i{i}^{\mathrm{\prime }}}$ and $C(w_j,w_{j^{\mathrm{\prime }}})=c_j{\delta }_{j{j}^{\mathrm{\prime }}}$ for real $b_i,c_j$. No $b_i$ is zero: a zero entry would make $v_i$ pair to zero with every basis vector, hence lie in the radical by [F5], contradicting nondegeneracy; the same holds for the $c_j$. After reordering, $b_i>0$ for $i\le p$, $b_i<0$ for $p<i\le n$, and $c_j>0$ for $j\le p^{\mathrm{\prime }}$, $c_j<0$ for $p^{\mathrm{\prime }}<j\le m$.

By [F6] the products $v_i\otimes w_j$ form a basis of $V{\otimes }_{\mathbb{R}}W$. Define a bilinear form $D$ on $V{\otimes }_{\mathbb{R}}W$ by prescribing its diagonal matrix in this basis, $D(v_i\otimes w_j,v_{i^{\mathrm{\prime }}}\otimes w_{j^{\mathrm{\prime }}}):=b_i{c}_j{\delta }_{i{i}^{\mathrm{\prime }}}{\delta }_{j{j}^{\mathrm{\prime }}}$, and extending by the matrix formula of [F5]: this is a well-defined bilinear form, and it is symmetric because the diagonal matrix is symmetric.

For $v={\sum }_i{a}_i{v}_i$, $w={\sum }_j{d}_j{w}_j$, $v^{\mathrm{\prime }}={\sum }_{i^{\mathrm{\prime }}}{a}_{i^{\mathrm{\prime }}}^{\mathrm{\prime }}{v}_{i^{\mathrm{\prime }}}$, $w^{\mathrm{\prime }}={\sum }_{j^{\mathrm{\prime }}}{d}_{j^{\mathrm{\prime }}}^{\mathrm{\prime }}{w}_{j^{\mathrm{\prime }}}$ in $V$ and $W$, bilinearity and the diagonal prescription give $D(v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }})={\sum }_{i,i^{\mathrm{\prime }},j,j^{\mathrm{\prime }}}{a}_i{d}_j{a}_{i^{\mathrm{\prime }}}^{\mathrm{\prime }}{d}_{j^{\mathrm{\prime }}}^{\mathrm{\prime }}{b}_i{c}_j{\delta }_{i{i}^{\mathrm{\prime }}}{\delta }_{j{j}^{\mathrm{\prime }}}=\left({\sum }_i{a}_i{a}_i^{\mathrm{\prime }}{b}_i\right)\left({\sum }_j{d}_j{d}_j^{\mathrm{\prime }}{c}_j\right)=B(v,v^{\mathrm{\prime }})C(w,w^{\mathrm{\prime }})$. Hence $D$ is exactly the form $B\otimes C$ of the statement on simple tensors, and by [F7] it is the unique bilinear form with these values on elementary tensors.

Inertia: the basis $(v_i\otimes w_j)$ is diagonal for $D$ with entries $b_i{c}_j$, none zero. The entry $b_i{c}_j$ is positive exactly when $b_i$ and $c_j$ have the same sign, which happens for $p{p}^{\mathrm{\prime }}$ pairs of positive entries and $q{q}^{\mathrm{\prime }}$ pairs of negative entries; it is negative for the $p{q}^{\mathrm{\prime }}$ pairs with $b_i>0>c_j$ and the $p^{\mathrm{\prime }}q$ pairs with $c_j>0>b_i$. By [F1] the inertia of $D$ is therefore $(p{p}^{\mathrm{\prime }}+q{q}^{\mathrm{\prime }},p{q}^{\mathrm{\prime }}+p^{\mathrm{\prime }}q,0)$.

Nondegeneracy: let $u={\sum }_{i,j}{x}_{ij}{v}_i\otimes w_j$ satisfy $D(u,u^{\mathrm{\prime }})=0$ for all $u^{\mathrm{\prime }}$. Evaluating on $u^{\mathrm{\prime }}=v_i\otimes w_j$ and using the diagonal prescription gives $x_{ij}{b}_i{c}_j=0$, and $b_i{c}_j\mathrm{\ne }0$, so every $x_{ij}=0$ and $u=0$. By symmetry the same computation with the second variable shows the right radical vanishes, so $D$ is nondegenerate.

Consequently $\mathrm{sign}(B\otimes C)=(p{p}^{\mathrm{\prime }}+q{q}^{\mathrm{\prime }})-(p{q}^{\mathrm{\prime }}+p^{\mathrm{\prime }}q)=(p-q)(p^{\mathrm{\prime }}-q^{\mathrm{\prime }})=\mathrm{sign}(B)\mathrm{sign}(C)$; if $V=0$ then $p=q=0$, and if $W=0$ then $p^{\mathrm{\prime }}=q^{\mathrm{\prime }}=0$; in either case the tensor-product basis is empty and both sides vanish. By [F8] the inertia computation pins down the congruence class of $D$, and [F2] gives its normal form.