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Congruent matrices have the same rank; hence rank and nondegeneracy of a bilinear form are basis-independent
Statement
If with invertible, then . Consequently the rank and nondegeneracy of a bilinear form do not depend on the basis used to represent it.
Facts & Assumptions
Given: Congruent matrices with invertible.
Congruence is precisely the matrix relation arising from a basis change for a bilinear form (A basis change by changes the matrix of a bilinear form from to ).
Transpose reverses products and sends an inverse to the inverse transpose (Transpose is linear and involutive, and ).
An invertible linear map is a linear isomorphism and has a two-sided linear inverse (Invertible linear maps, linear isomorphisms, and inverse linear maps). Two finite-dimensional spaces are linearly isomorphic if and only if they have the same dimension (Two finite-dimensional vector spaces over are linearly isomorphic if and only if they have the same dimension).
Proof
Right multiplication by the invertible is precomposition with an isomorphism, so . Left multiplication by maps this image isomorphically to , because [L3] makes invertible.
Therefore the two image spaces have the same dimension, so .
By [L1], matrices of one bilinear form in two bases are congruent. Rank is thus basis-independent, and nondegeneracy is also basis-independent because it is equivalent to invertibility of a representing matrix, which the congruence relation preserves in both directions.
Depends on
- A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{\mathsf T}AP$
- $[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}$
- Transpose is linear and involutive, and $(AB)^{\mathsf T}=B^{\mathsf T}A^{\mathsf T}$
- Invertible linear maps, linear isomorphisms, and inverse linear maps
- Two finite-dimensional vector spaces over $F$ are linearly isomorphic if and only if they have the same dimension
Used by
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Sources
- H. Pinkham, Linear Algebra, §7.7 (standard reference, not scraped)