CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Determinant is a group homomorphism $GL (V) \to F^{\times}$ , and $\det (T^{- 1}) = \det (T)^{- 1}$

Statement

For a finite-dimensional vector space $V$ over a field $F$ , determinant restricts to a group homomorphism

$\det : GL (V) \to F^{\times} .$

For every $T \in GL (V)$ , $\det (T^{- 1}) = \det (T)^{- 1}$ .

Facts & Assumptions

Given: $V, F$ , and invertible operators on $V$ .

[L1]

$\det (S \circ T) = \det (S) \det (T)$ (For endomorphisms $S$ and $T$ of one finite-dimensional vector space, $\det (S T) = \det (S) \det (T)$ ).

[L2]

An operator is invertible exactly when its determinant is nonzero (A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).

[F1]

An invertible linear map has a two-sided inverse (Invertible linear maps, linear isomorphisms, and inverse linear maps).

[L3]

The units of a commutative ring form a group (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).

Proof

technique · direct

1.1L2L3

If $T$ is invertible, [L2] gives $\det (T) \neq 0$ , so $\det (T) \in F^{\times}$ and [L3] supplies its inverse.

1.2L1L2algebra

Put $d = \det (I_{V})$ . By [L1], $d = d^{2}$ , and [L2] gives $d \neq 0$ because $I_{V}$ is invertible. Field cancellation yields $d = 1$ .

2.1L1step 1.2

Multiplicativity [L1] and step 1.2 show that determinant preserves the group product and identity.

2.2F1L1step 1.2L3

For invertible $T$ , [F1] gives $T \circ T^{- 1} = I_{V}$ . Applying [L1] and step 1.2 yields $\det (T) \det (T^{- 1}) = 1$ , so uniqueness of inverses in [L3] gives $\det (T^{- 1}) = \det (T)^{- 1}$ .

3.1step 1.1step 2.1step 2.2∎

Steps 1.1, 2.1, and 2.2 prove the homomorphism and inverse claims.

Depends on

Dependency tree · 2 levels

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed.

Determinant is a group homomorphism GL⁡(V)→F×GL(V)→F×, and det⁡(T−1)=det⁡(T)−1det(T−1)=det(T)−1