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A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero
Statement
Let be a linear operator on a finite-dimensional vector space over a field . Then is invertible if and only if .
Facts & Assumptions
Given: as in the statement.
If is invertible, its representing matrix is invertible (A square matrix is invertible exactly when its multiplication map is a linear isomorphism; matrices preserve inverses of linear isomorphisms).
Every square matrix is the representing matrix of a unique operator, and composition of operators becomes multiplication of their matrices ( is a vector-space isomorphism , ).
A positive-sized square matrix over a commutative ring is invertible exactly when its determinant is a unit (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).
A field is a commutative ring and its units are precisely its nonzero elements (Every field is a commutative ring with ; it is an integral domain, and it is a commutative division ring).
The operator determinant is the determinant of a representing matrix and is basis independent (The determinant of a linear operator is independent of the chosen ordered basis).
In dimension zero, the operator determinant is defined to be (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and on the zero space).
Proof
Suppose and choose an ordered basis, writing . If is invertible, then is invertible by [L1]. Conversely, if is invertible, [L4] gives an operator with ; the composite formula gives , and injectivity of matrix representation gives . Thus is invertible exactly when is invertible.
If , the unique endomorphism is the identity and is invertible, while [F2] gives .
By [L2] and [F1], is invertible exactly when ; by [L3], this is exactly .
Steps 2.1 and 1.2 establish the equivalence in every finite dimension.
Depends on
- The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space
- The determinant of a linear operator is independent of the chosen ordered basis
- A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit
- A square matrix is invertible exactly when its multiplication map is a linear isomorphism; matrices preserve inverses of linear isomorphisms
- $T\mapsto[T]_{\mathcal B}^{\mathcal C}$ is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$
- $[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}$
- Every field is a commutative ring with $1 \ne 0$; it is an integral domain, and it is a commutative division ring
Used by
- Determinant is a group homomorphism GL(V)→ F^×, and det(T⁻¹)=det(T)⁻¹ Corollary
- The inverse of an invertible finite-dimensional endomorphism is a polynomial in that endomorphism Corollary
- A nonidentity projection of the plane has determinant zero and is not invertible Example
- Projection onto a finite-dimensional subspace by a Gram matrix Example
- The quaternion double cover generates the third homotopy group of SO(3) Lemma
- A Gram determinant is nonnegative and is positive exactly when the vector list is linearly independent Theorem
- Every invertible real or complex square matrix has a unique factorisation A=QR with Q orthogonal or unitary and R upper triangular with positive real diagonal Theorem
- For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and T^*T=I are equivalent Theorem
- For every finite-dimensional space, σ_F(T) is exactly the set of roots in F of χ_T Theorem
Dependency tree · two levels
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Sources
- Sheldon Axler, Linear Algebra Done Right, 4th ed. (standard reference, not scraped)