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A basis change by changes the matrix of a bilinear form from to
Statement
Let be a bilinear form on a finite-dimensional space. If is its matrix in an old basis and the columns of an invertible matrix are the new basis vectors in old coordinates, then its matrix in the new basis is
Matrices related by with invertible are called congruent.
Facts & Assumptions
Given: The form, the two bases, and the change-of-basis matrix described above.
If are coordinate columns, the matrix of satisfies (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).
Transpose reverses products: (Transpose is linear and involutive, and ).
Matrix multiplication is associative and represents the relevant finite sums (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).
Proof
If are the new coordinate columns of , respectively, [L2] makes their old coordinate columns .
By [L1], , using [L3] and [L4].
Since step 2.1 holds for every , the new matrix is . The displayed relation is therefore exactly the equivalence generated by basis changes and is called congruence, not similarity.
Depends on
- The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space
- $[T(v)]_{\mathcal C}=[T]_{\mathcal B}^{\mathcal C}[v]_{\mathcal B}$
- Transpose is linear and involutive, and $(AB)^{\mathsf T}=B^{\mathsf T}A^{\mathsf T}$
- Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication
Used by
- Congruent matrices have the same rank; hence rank and nondegeneracy of a bilinear form are basis-independent Corollary
- Over a field of characteristic not 2, every symmetric matrix is congruent to a diagonal matrix Corollary
- Two real symmetric bilinear forms are congruent if and only if they have the same inertia Corollary
- Congruence need not preserve trace or determinant: the real 1×1 matrices [1] and [4] are congruent Counterexample
- For symmetric M=beginpmatrixA&BB^T&C endpmatrix with A invertible, a block-unitriangular congruence gives A⊕(C-B^TA⁻¹B) and factors det M Lemma
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Sources
- H. Pinkham, Linear Algebra, §7.7 (standard reference, not scraped)