Linear Algebra 31 | Inverses of Linear Maps are Linear [dark version]

TL;DR

If F: R^n → R^N is linear and bijective, its inverse map F^{-1} exists and is well-defined.

Briefing Cornell Notes

Briefing

Inverses of bijective linear maps are themselves linear—so once a linear map is invertible, there’s no need to re-check linearity for its inverse. The key takeaway matters because it lets students move confidently between a linear transformation and its inverse when solving systems, changing coordinates, or manipulating matrix equations.

The discussion starts with the familiar matrix setting: if a linear map is represented by a matrix, then the inverse map corresponds to the inverse matrix. When two linear maps are composed, their matrices multiply in the same order as the maps. That composition has an inverse too, and the inverse of a product reverses the order: the inverse of AB is B^{-1}A^{-1}. This order-reversal is presented as the concrete rule that mirrors how inverses behave under composition.

From there, the focus shifts to an abstract (non-matrix) proof. Consider a linear map F from R^n to R^N that is bijective, ensuring that F^{-1} exists as a well-defined function from R^N back to R^n. To prove F^{-1} is linear, the argument checks two defining properties: preservation of scalar multiplication and preservation of addition.

For scalar multiplication, take any vector Y in the codomain and a scalar Λ. The proof begins with F^{-1}(ΛY). Using bijectivity, there exists a unique X such that F(X)=Y. Linearity of F then allows Λ to be pulled inside: F(ΛX)=ΛF(X)=ΛY. Applying F^{-1} cancels F, leaving ΛX, which is exactly ΛF^{-1}(Y). So F^{-1}(ΛY)=ΛF^{-1}(Y), establishing scalar compatibility.

For addition, take two vectors Y and Ỹ in the codomain. Again, bijectivity provides vectors X and X̃ in the domain such that F(X)=Y and F(X̃)=Ỹ. Linearity of F gives F(X+X̃)=F(X)+F(X̃)=Y+Ỹ. Applying F^{-1} to both sides cancels F and yields F^{-1}(Y+Ỹ)=X+X̃=F^{-1}(Y)+F^{-1}(Ỹ). This confirms additivity.

With both scalar multiplication and addition preserved, F^{-1} is linear. The result is framed as a practical rule: if a map is linear and bijective, its inverse automatically inherits linearity, eliminating redundant verification steps in future problems.

Cornell Notes

A bijective linear map F has an inverse function F^{-1} that is guaranteed to be linear. The proof checks the two linearity requirements directly: F^{-1}(ΛY)=ΛF^{-1}(Y) for any scalar Λ and vector Y, and F^{-1}(Y+Ỹ)=F^{-1}(Y)+F^{-1}(Ỹ) for any vectors Y and Ỹ. Bijectivity is what makes F^{-1} well-defined and ensures unique preimages X and X̃ exist. Linearity of F is then used to move scalars and sums through F, after which applying F^{-1} cancels F. This mirrors the matrix fact that (AB)^{-1}=B^{-1}A^{-1} and extends it to abstract linear maps.

Why does bijectivity matter for proving that the inverse map exists and is usable?

Bijectivity ensures that for every Y in the codomain there is exactly one X in the domain with F(X)=Y. That uniqueness is what makes F^{-1}(Y) well-defined. Without bijectivity, F^{-1} might not exist everywhere or might not be single-valued, so expressions like F^{-1}(ΛY) and F^{-1}(Y+Ỹ) wouldn’t be reliably defined.

How does the proof establish that F^{-1} preserves scalar multiplication?

Start with F^{-1}(ΛY). By bijectivity, pick the unique X such that F(X)=Y, so F^{-1}(Y)=X. Linearity of F gives F(ΛX)=ΛF(X)=ΛY. Applying F^{-1} cancels F, yielding F^{-1}(ΛY)=ΛX=ΛF^{-1}(Y).

How does the proof establish that F^{-1} preserves addition?

Take vectors Y and Ỹ and use bijectivity to choose unique X and X̃ such that F(X)=Y and F(X̃)=Ỹ. Linearity of F implies F(X+X̃)=F(X)+F(X̃)=Y+Ỹ. Applying F^{-1} cancels F, giving F^{-1}(Y+Ỹ)=X+X̃=F^{-1}(Y)+F^{-1}(Ỹ).

What matrix rule about inverses is used as motivation for the abstract result?

In the matrix setting, composing linear maps corresponds to multiplying matrices. The inverse of a product reverses order: (AB)^{-1}=B^{-1}A^{-1}. This order reversal reflects how inverses interact with composition and motivates the general behavior of inverse maps.

What is the practical takeaway for students working with invertible linear maps?

If F is linear and bijective, then F^{-1} is automatically linear. That means there’s no need to separately verify linearity of the inverse transformation when solving equations or transforming between vector spaces.

Review Questions

What two properties must be checked to prove a function is linear, and how are they verified for F^{-1}?
Where exactly does bijectivity enter the proof, and what would fail without it?
How does the abstract proof’s cancellation step (applying F^{-1} to an equation involving F) work conceptually?

Key Points

1
If F: R^n → R^N is linear and bijective, its inverse map F^{-1} exists and is well-defined.
2
For any scalar Λ and vector Y, the inverse satisfies F^{-1}(ΛY)=ΛF^{-1}(Y).
3
For any vectors Y and Ỹ, the inverse satisfies F^{-1}(Y+Ỹ)=F^{-1}(Y)+F^{-1}(Ỹ).
4
The proof relies on bijectivity to guarantee unique preimages X and X̃ with F(X)=Y and F(X̃)=Ỹ.
5
Linearity of F is used to move scalars and sums inside F before applying F^{-1} to cancel F.
6
In the matrix representation, the inverse of a product reverses order: (AB)^{-1}=B^{-1}A^{-1}.
7
Once a linear map is invertible, linearity of the inverse comes for free—no separate linearity check is needed.

Highlights

A bijective linear map’s inverse automatically preserves addition and scalar multiplication.

The scalar-multiplication proof uses the identity F(ΛX)=ΛF(X) and then cancels with F^{-1}.

The addition proof uses F(X+X̃)=F(X)+F(X̃) and then cancels with F^{-1}.

Matrix intuition matches the abstract result: inverses reverse order under composition, (AB)^{-1}=B^{-1}A^{-1}. 

Topics

Inverse Linear Maps
Linearity
Bijective Transformations
Matrix Inverses
Composition of Linear Maps