Composition of Invertible Linear Maps: A Proof

Proving the Invertibility of Composed Linear Maps

This proof demonstrates that the composition of two invertible linear maps is itself invertible.

Theorem: If S and T are invertible linear maps, then their composition, ST, is also invertible.

Proof:

Let's assume:

• S ∈ L(V,W), meaning S is a linear map from vector space V to W.* T ∈ L(U,V), meaning T is a linear map from vector space U to V.

The map ST represents the composition of S and T.

To prove ST is invertible, we need to show that if S and T are individually invertible, then ST also has an inverse.

1. Inverses of S and T:

   • Let S^(-1) denote the inverse of S. By definition, S^(-1) is a linear map from W to V such that: * S^(-1) o S = I_V (Identity map on V) * S o S^(-1) = I_W (Identity map on W)

   • Similarly, let T^(-1) represent the inverse of T. T^(-1) is a linear map from V to U satisfying: * T^(-1) o T = I_U (Identity map on U) * T o T^(-1) = I_V (Identity map on V)

2. Composition and Inverses:

   Now, let's examine the composition (S o T) and its potential inverse (T^(-1) o S^(-1)):

   • Left Inverse: (S o T) o (T^(-1) o S^(-1)) = S o (T o T^(-1)) o S^(-1) (Associativity of composition) Since T o T^(-1) = I_V, we get: (S o T) o (T^(-1) o S^(-1)) = S o I_V o S^(-1) = S o S^(-1) = I_W This demonstrates that (T^(-1) o S^(-1)) acts as a left inverse for (S o T).

   • Right Inverse: Similarly, we can show (T^(-1) o S^(-1)) is also the right inverse of (S o T): (T^(-1) o S^(-1)) o (S o T) = T^(-1) o (S^(-1) o S) o T = T^(-1) o I_V o T = T^(-1) o T = I_U

3. Conclusion:

Since (S o T) has both a left inverse and a right inverse, and they are equal, we can conclude that (S o T) is invertible, and its inverse is (T^(-1) o S^(-1)).

Therefore, we have successfully proven that the composition of two invertible linear maps is indeed invertible.