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A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit
Statement
Let be a commutative ring, , and . Then is invertible if and only if is a unit of .
Facts & Assumptions
Given: as in the statement.
If a positive-sized square matrix is invertible, its determinant is a unit (An invertible square matrix over a commutative ring has unit determinant).
A square matrix is invertible when it has a two-sided multiplicative inverse (Invertible square matrices and similarity over a commutative ring).
A unit has a unique multiplicative inverse (The units of a ring are the invertible elements of its multiplicative monoid, and is a group under multiplication; only in the zero ring).
Proof
If is invertible, [L1] says directly that is a unit.
Conversely, suppose is a unit, and let be its inverse.
Multiplying both identities in [L2] by the scalar gives .
Thus is a two-sided inverse, so is invertible. Together with step 1.1, this proves both directions.
Depends on
- For every positive-sized square matrix over a commutative ring, $A\operatorname{adj}(A)=\operatorname{adj}(A)A=\det(A)I$
- An invertible square matrix over a commutative ring has unit determinant
- Invertible square matrices and similarity over a commutative ring
- The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring
Used by
- If det(A) is a unit, then A⁻¹=det(A)⁻¹adj(A) Corollary
- Over a field, for each right-hand side b, Ax=b has a unique solution exactly when det(A) is nonzero, and then Cramer's quotient formula holds Corollary
- The index of a full-rank subgroup of ℤⁿ is the absolute determinant of a generating matrix Corollary
- Multiplication by 2 on ℤ is injective but not surjective: its determinant is the non-unit 2, its adjugate is integral, and its inverse exists after extending scalars to ℚ Example
- Unitary and special unitary Lie groups Example
- FALSE: a square matrix over a commutative ring is invertible if and only if its determinant is nonzero False statement
- For a finite Galois extension, (αⱼ) is a base-field basis exactly when the matrix (σᵢαⱼ) is invertible Lemma
- A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero Theorem
- Cramer's rule over a commutative ring: every solution satisfies det(A)xⱼ=det(Aⱼ(b)), and a unit determinant gives the unique quotient formula Theorem
- Every finite Galois extension of an infinite field has a normal basis Theorem
Dependency tree · two levels
18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- András Pál, Introduction to Commutative Algebra, Proposition 7.2 (standard reference, not scraped)