 Determinant pairing is not an inner product

On real symmetric matrices of size $n\ge 2$, the displayed pairing is symmetric because $\det (AC)=\det A\det C$. It is not bilinear: $B(I,2I)=2^n$, whereas linearity would require $2B(I,I)=2$. Its self-pairing is always nonnegative, but a nonzero singular matrix has $B(A,A)=0$, so positive definiteness fails as well. For $n=1$ it reduces to the ordinary real inner product. The Frobenius inner product $\mathrm{tr}(A^{T}C)$ supplies the appropriate bilinear, positive-definite geometry in every size.