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For a real symmetric bilinear form, a basis puts its quadratic expression in diagonal form with $p$ positive, $q$ negative and $r$ zero squares. Its rank is $p+q$, the rank of the coefficient matrix, or equivalently the codimension of its radical. Its signature is $p-q$; the signature pair is $(p,q)$, another common convention.

These quantities do not depend on the basis. A basis change replaces the matrix $M$ by $P^{T}MP$ with $P$ invertible, preserving matrix rank. The number $p$ is intrinsically the largest dimension of a subspace on which the form is positive definite. In a diagonal representation the positive coordinate space attains this dimension; any subspace of dimension greater than $p$ intersects the negative-plus-null coordinate space nontrivially and therefore cannot be positive definite. The same argument with signs reversed characterizes $q$. This proves the needed invariance, rather than assuming it from a chosen representation.

Decompose the full matrix space asusing the symmetric traceless, skew-symmetric and scalar parts. The summands are orthogonal for the given form. Indeed $\mathrm{tr}(SK)=0$ for symmetric $S$ and skew-symmetric $K$, by transposition and cyclic invariance of trace, while both non-scalar summands are traceless and hence orthogonal to $I$. On them,for nonzero vectors in their respective summands. The dimensions are $n(n+1)/2-1$, $n(n-1)/2$ and one. Therefore the modified trace form on real matrices hasThere is no radical for $n\ge 2$.