Page 68 - Linear Algebra Done Right

Chapter 3. Linear Maps
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S = SI = S(TS ) = (ST)S = IS = S ,

so S = S . In other words, if T is invertible, then it has a unique

inverse, which we denote by T −1 . Rephrasing all this once more, if
T ∈L(V, W) is invertible, then T −1 is the unique element of L(W, V)
such that T −1 T = I and TT −1 = I. The following proposition charac-
terizes the invertible linear maps.
3.17 Proposition: A linear map is invertible if and only if it is injec-
tive and surjective.
Proof: Suppose T ∈L(V, W). We need to show that T is invertible
if and only if it is injective and surjective.
First suppose that T is invertible. To show that T is injective, sup-
pose that u, v ∈ V and Tu = Tv. Then

u = T −1 (Tu) = T −1 (Tv) = v,

so u = v. Hence T is injective.
We are still assuming that T is invertible. Now we want to prove
that T is surjective. To do this, let w ∈ W. Then w = T(T −1 w), which
shows that w is in the range of T. Thus range T = W, and hence T is
surjective, completing this direction of the proof.
Now suppose that T is injective and surjective. We want to prove
that T is invertible. For each w ∈ W, define Sw to be the unique ele-
ment of V such that T(Sw) = w (the existence and uniqueness of such
an element follow from the surjectivity and injectivity of T). Clearly
TS equals the identity map on W. To prove that ST equals the identity
map on V, let v ∈ V. Then

T(STv) = (TS)(Tv) = I(Tv) = Tv.

This equation implies that STv = v (because T is injective), and thus
ST equals the identity map on V. To complete the proof, we need to
show that S is linear. To do this, let w 1 , w 2 ∈ W. Then

T(Sw 1 + Sw 2 ) = T(Sw 1 ) + T(Sw 2 ) = w 1 + w 2 .
Thus Sw 1 +Sw 2 is the unique element of V that T maps to w 1 +w 2 .By
the definition of S, this implies that S(w 1 + w 2 ) = Sw 1 + Sw 2 . Hence
S satisfies the additive property required for linearity. The proof of
homogeneity is similar. Specifically, if w ∈ W and a ∈ F, then