Composition and Inversion of Isomorphisms

Isomorphisms Can Be Composed and Reversed

An isomorphism is a linear map that gives a one-to-one correspondence between two vector spaces. When one isomorphism carries vectors from \(U\) to \(V\), and another carries them from \(V\) to \(W\), applying the maps in sequence gives a new correspondence from \(U\) to \(W\). An isomorphism can also be run in reverse: its inverse map takes each output back to the unique input that produced it. These operations are useful because they let us combine known correspondences and calculate new ones without testing every vector separately.

Throughout, \(U,V,W\) are vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A linear map \(T:U\to V\) preserves addition and scalar multiplication: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(c\mathbf{u})=cT(\mathbf{u})\). It is bijective if it is both one-to-one (distinct inputs have distinct outputs) and onto (every element of the codomain is an output).

The inverse function exists because \(T\) is bijective: being onto supplies an input for every output, and being one-to-one makes that input unique. Linearity of the inverse is an additional fact, not part of the definition of an inverse function. We prove it first, then use it to establish the composition and inversion rules.

Proof. The inverse function is bijective: \(T^{-1}(\mathbf{y}_1)=T^{-1}(\mathbf{y}_2)\) implies, after applying \(T\), that \(\mathbf{y}_1=\mathbf{y}_2\); and for each \(\mathbf{x}\in U\), \(T^{-1}(T(\mathbf{x}))=\mathbf{x}\), so \(\mathbf{x}\) is in the image of \(T^{-1}\). To prove linearity, take \(\mathbf{y}_1,\mathbf{y}_2\in V\) and scalars \(a,b\in\mathbb{F}\). Set \(\mathbf{x}_1=T^{-1}(\mathbf{y}_1)\) and \(\mathbf{x}_2=T^{-1}(\mathbf{y}_2)\). By linearity of \(T\), \(T(a\mathbf{x}_1+b\mathbf{x}_2)=a\mathbf{y}_1+b\mathbf{y}_2\). The unique input that \(T\) sends to \(a\mathbf{y}_1+b\mathbf{y}_2\) is therefore \(a\mathbf{x}_1+b\mathbf{x}_2\). By the definition of \(T^{-1}\), this gives \(T^{-1}(a\mathbf{y}_1+b\mathbf{y}_2)=aT^{-1}(\mathbf{y}_1)+bT^{-1}(\mathbf{y}_2)\). Thus \(T^{-1}\) is linear as well as bijective, so it is an isomorphism. \(\square\)

The Composition Theorem

Suppose \(T:U\to V\) and \(S:V\to W\) are isomorphisms. The composition \(S\circ T:U\to W\) means that \(T\) is applied first and then \(S\): \((S\circ T)(\mathbf{u})=S(T(\mathbf{u}))\). Its inverse must undo these steps in the opposite order. The next theorem proves both that the composition is an isomorphism and that its inverse has this reverse-order form.

Proof. For \(\mathbf{u}_1,\mathbf{u}_2\in U\) and \(a,b\in\mathbb{F}\), linearity of \(T\) and \(S\) gives \((S\circ T)(a\mathbf{u}_1+b\mathbf{u}_2)=S(aT(\mathbf{u}_1)+bT(\mathbf{u}_2))=a(S\circ T)(\mathbf{u}_1)+b(S\circ T)(\mathbf{u}_2)\). Hence the composition is linear. If \((S\circ T)(\mathbf{u}_1)=(S\circ T)(\mathbf{u}_2)\), injectivity of \(S\) gives \(T(\mathbf{u}_1)=T(\mathbf{u}_2)\), and injectivity of \(T\) then gives \(\mathbf{u}_1=\mathbf{u}_2\). So the composition is one-to-one. For any \(\mathbf{w}\in W\), surjectivity of \(S\) gives \(\mathbf{v}\in V\) with \(S(\mathbf{v})=\mathbf{w}\); surjectivity of \(T\) gives \(\mathbf{u}\in U\) with \(T(\mathbf{u})=\mathbf{v}\). Thus \((S\circ T)(\mathbf{u})=\mathbf{w}\), so the composition is onto. It is therefore an isomorphism.

To verify the inverse formula, start with any \(\mathbf{w}\in W\). Applying \(S^{-1}\) and then \(T^{-1}\) returns an element of \(U\), and \((S\circ T)(T^{-1}(S^{-1}(\mathbf{w})))=S(S^{-1}(\mathbf{w}))=\mathbf{w}\). In the other direction, for any \(\mathbf{u}\in U\), \(T^{-1}(S^{-1}((S\circ T)(\mathbf{u})))=T^{-1}(T(\mathbf{u}))=\mathbf{u}\). These two identities show that \(T^{-1}\circ S^{-1}\) is the inverse of \(S\circ T\). \(\square\)

The order is important. In \(S\circ T\), the map \(T\) acts first, so its inverse acts last when undoing the composition. The formula also has the correct domains: \(S^{-1}:W\to V\) acts first on an element of \(W\), followed by \(T^{-1}:V\to U\).

Computing a Composition with Matrices

An Isomorphism on a Matrix Space

The same reasoning applies to vector spaces whose elements are matrices. For this example, \(\mathbb{R}^{2\times2}\) is a vector space with matrix addition and scalar multiplication. We will specify a map by its effect on the entries of a matrix, then find its inverse by solving for the original entries.

Computing the Reverse Order Directly

Why the Order and Hypotheses Matter

The composition theorem requires the codomain of the first map to be the domain of the second. It also requires both maps to be isomorphisms. Linearity alone does not guarantee that a map has an inverse: for example, the zero map from \(\mathbb{R}^2\) to \(\mathbb{R}^2\) sends every vector to \(\mathbf{0}\), so it is neither one-to-one nor onto. The inverse rule is for isomorphisms, not for arbitrary linear maps.

Common mistake: reversing the wrong order. For the first worked example, the correct inverse of \(S\circ T\) has matrix \(\mathbf{A}^{-1}\mathbf{B}^{-1}\). A tempting but incorrect expression is \(\mathbf{B}^{-1}\mathbf{A}^{-1}\). In fact, that reversed product is

It is not an inverse of \(\mathbf{B}\mathbf{A}\), since multiplying gives \(\begin{bmatrix}2&4\\1&3\end{bmatrix}\begin{bmatrix}\tfrac12&-1\\-\tfrac12&2\end{bmatrix} =\begin{bmatrix}-1&6\\-1&5\end{bmatrix}\), not the identity matrix. The correction is to undo the last map applied first: since \(T\) acts before \(S\), undo \(S\) first and then undo \(T\). That is exactly \(T^{-1}\circ S^{-1}\).

These rules are useful whenever a complicated correspondence is built from simpler ones. If each stage is an isomorphism, the full sequence is an isomorphism, and its inverse can be computed by reversing the stages and inverting each map. Checking the two compositions against the appropriate identity map is a direct way to catch both arithmetic errors and order errors.