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For the linear-first convention, a basis change by sends a sesquilinear matrix to ; Hermitian forms satisfy
Statement
Let be sesquilinear over a field with involution , using the convention linear in the first variable. If its old matrix is and a basis change has matrix , then its new matrix is
where is obtained entrywise. Moreover, is Hermitian if and only if .
Facts & Assumptions
Given: A field involution , a sesquilinear form , and the displayed basis data.
Sesquilinearity is linear in the first variable and -linear in the second; Hermitian symmetry is (Sesquilinear and Hermitian forms over a field with an involution, using the convention linear in the first variable).
Matrix multiplication expands as the corresponding finite row-column sums (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).
Transpose reverses products (Transpose is linear and involutive, and ).
Proof
If are new coordinate columns, their old columns are . Expanding [L1] in an old basis gives .
In a basis , Hermitian symmetry is for every , which is exactly .
Since step 1.1 holds for every , the new matrix is . This includes the identity involution, where it reduces to ordinary congruence.
Conversely, if , the coordinate formula and give for arbitrary coordinate columns, so is Hermitian.
Steps 2.1, 1.2, and 2.2 prove the basis-change formula and both directions of the Hermitian criterion.
Depends on
- Sesquilinear and Hermitian forms over a field with an involution, using the convention linear in the first variable
- Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication
- Transpose is linear and involutive, and $(AB)^{\mathsf T}=B^{\mathsf T}A^{\mathsf T}$
Used by
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Sources
- H. Pinkham, Linear Algebra, §7.8 (standard reference, not scraped)