The Inverse Map Needs Its Own Linearity Proof
An isomorphism is a linear map that is also bijective. Bijectivity gives the map an inverse function, but being an inverse function does not by itself guarantee linearity. To establish that the inverse is linear, we must check the two defining properties: it must preserve addition and scalar multiplication. The key is to represent vectors in its domain as images under the original map, then use the original map’s linearity and the uniqueness of preimages.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A map \(T:V\to W\) is linear if \(T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})\) and \(T(c\mathbf{x})=cT(\mathbf{x})\) for all \(\mathbf{x},\mathbf{y}\in V\) and \(c\in\mathbb{F}\). It is bijective if every element of \(W\) is the image of exactly one element of \(V\). No finite-dimensional assumption is needed for the result below.
There are two parts to the phrase “unique preimage.” Surjectivity guarantees that a preimage exists for every vector in \(W\); injectivity guarantees that two different vectors in \(V\) cannot both be preimages of the same vector. Both parts matter: without existence, the inverse would not be defined everywhere on \(W\), and without uniqueness, it would not assign a single output to each input.
Proof. Since \(T\) is surjective, for each \(\mathbf{y}\in W\) there is at least one \(\mathbf{x}\in V\) with \(T(\mathbf{x})=\mathbf{y}\). Suppose \(\mathbf{x}_1,\mathbf{x}_2\in V\) both satisfy \(T(\mathbf{x}_1)=\mathbf{y}\) and \(T(\mathbf{x}_2)=\mathbf{y}\). Then \(T(\mathbf{x}_1)=T(\mathbf{x}_2)\), so injectivity gives \(\mathbf{x}_1=\mathbf{x}_2\). Therefore the preimage exists and is unique. \(\square\)
Proving the Inverse Preserves Linear Combinations
Proof. Let \(\mathbf{u},\mathbf{v}\in W\). By bijectivity, there are unique \(\mathbf{x},\mathbf{y}\in V\) such that \(T(\mathbf{x})=\mathbf{u}\) and \(T(\mathbf{y})=\mathbf{v}\). Linearity of \(T\) gives
So \(\mathbf{x}+\mathbf{y}\) is a preimage of \(\mathbf{u}+\mathbf{v}\). The preimage is unique, hence \(T^{-1}(\mathbf{u}+\mathbf{v})=\mathbf{x}+\mathbf{y}=T^{-1}(\mathbf{u})+T^{-1}(\mathbf{v})\). Now let \(c\in\mathbb{F}\). Linearity of \(T\) also gives
Thus \(c\mathbf{x}\) is a preimage of \(c\mathbf{u}\); uniqueness gives \(T^{-1}(c\mathbf{u})=c\mathbf{x}=cT^{-1}(\mathbf{u})\). These equalities hold for all \(\mathbf{u},\mathbf{v}\in W\) and \(c\in\mathbb{F}\), so \(T^{-1}\) preserves addition and scalar multiplication and is linear. \(\square\)
The proof does not assume the conclusion. In particular, it does not begin by applying \(T^{-1}\) to a sum and distributing it. Instead, it finds an explicit preimage of the sum using linearity of \(T\), then uses uniqueness to identify that preimage with the value of \(T^{-1}\). The same reasoning applies to scalar multiples. When \(c=0\), the argument still works: \(T(0\mathbf{x})=T(\mathbf{0})=\mathbf{0}=0\mathbf{u}\), so uniqueness identifies the preimage as \(\mathbf{0}\).
Worked Examples with Preimages
Worked Example: A Linear Map from \(\mathbb{R}^2\) to \(\mathbb{R}^2\)
Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \(T(x_1,x_2)=(x_1+2x_2,3x_1+5x_2)\). Find its inverse and check the inverse formulas.
Write an output vector as \(\mathbf{y}=(y_1,y_2)\). To find its preimage, solve \(y_1=x_1+2x_2\) and \(y_2=3x_1+5x_2\). Subtracting the second equation from three times the first gives \(3y_1-y_2=x_2\). Then \(x_1=y_1-2x_2=y_1-2(3y_1-y_2)=-5y_1+2y_2\). Therefore the only possible inverse formula is
For every \((y_1,y_2)\), this formula gives a preimage. Indeed, applying \(T\) to it gives first coordinate \((-5y_1+2y_2)+2(3y_1-y_2)=y_1\), and second coordinate \(3(-5y_1+2y_2)+5(3y_1-y_2)=y_2\). The equations used to solve for \(x_1,x_2\) also show that no other preimage is possible. Thus the map is bijective and the displayed formula is its inverse.
For a numerical check, \(T(1,2)=(1+4,3+10)=(5,13)\), and \(T^{-1}(5,13)=(-25+26,15-13)=(1,2)\). For addition, take \(\mathbf{u}=(1,2)\) and \(\mathbf{v}=(3,-1)\). Then \(T^{-1}(\mathbf{u})=(-1,1)\), \(T^{-1}(\mathbf{v})=(-17,10)\), and
For scalar multiplication by \(2\), \(T^{-1}(2\mathbf{u})=T^{-1}(2,4)=(-2,2)=2(-1,1)=2T^{-1}(\mathbf{u})\). These computations check particular inputs; the theorem proves preservation for every input.
Worked Example: A Map from a Polynomial Space
Let \(P_1\) be the real vector space of polynomials of degree at most \(1\), including the zero polynomial. Define \(T:P_1\to\mathbb{R}^2\) by \(T(a+bt)=(a+b,a-b)\). Find the inverse and check how it handles a sum and a scalar multiple.
Given \((u,v)\in\mathbb{R}^2\), the preimage coefficients must satisfy \(a+b=u\) and \(a-b=v\). Adding these equations gives \(2a=u+v\), so \(a=(u+v)/2\). Subtracting the second equation from the first gives \(2b=u-v\), so \(b=(u-v)/2\). Thus
Substitution verifies the formula in both directions. Applying \(T\) to the proposed preimage gives coordinates \((\frac{u+v}{2}+\frac{u-v}{2},\frac{u+v}{2}-\frac{u-v}{2})=(u,v)\). Conversely, starting with \(a+bt\), the inverse applied to \(T(a+bt)=(a+b,a-b)\) has constant coefficient \(((a+b)+(a-b))/2=a\) and \(t\)-coefficient \(((a+b)-(a-b))/2=b\). The formulas therefore establish existence and uniqueness of preimages.
For a concrete addition check, let \(p=1+2t\) and \(q=3-t\). Then \(T(p)=(3,-1)\), \(T(q)=(2,4)\), \(T^{-1}(3,-1)=1+2t\), and \(T^{-1}(2,4)=3-t\). Since \(p+q=4+t\),
For scalar multiplication by \(-2\), \(T^{-1}(-2T(p))=T^{-1}(-6,2)=-2-4t=-2p\). This example illustrates the preimage argument in spaces whose vectors are polynomials rather than coordinate columns.
Why Both Hypotheses Matter
Linearity and bijectivity play different roles. Linearity of \(T\) ensures that sums and scalar multiples of preimages map to the corresponding sums and scalar multiples in \(W\). Bijectivity ensures that these candidates are the preimages selected by \(T^{-1}\): surjectivity supplies preimages for all inputs, while injectivity makes them unique. If either role is missing, the proof cannot be completed as stated.
Worked Example: A Bijection Whose Inverse Is Not Linear
Consider \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=x+1\). This function is bijective: for each \(y\in\mathbb{R}\), the unique solution of \(y=x+1\) is \(x=y-1\). Its inverse is \(g(y)=y-1\). But \(f\) is not linear, since \(f(0)=1\ne0\), and its inverse is not linear either: \(g(0+0)=-1\), whereas \(g(0)+g(0)=-2\). Thus bijectivity alone is not enough to guarantee a linear inverse.
The calculation also shows why the theorem cannot be used here: its hypothesis that the original map is linear fails. For contrast, in the polynomial example the map was both linear and bijective, and the inverse formula preserved addition and scalar multiplication.
A Common Mistake: Distributing the Inverse Without Proof
A circular argument might write \(T^{-1}(\mathbf{u}+\mathbf{v})=T^{-1}(\mathbf{u})+T^{-1}(\mathbf{v})\) and say that this follows from the definition of inverse. It does not follow from that definition alone: the definition only says which vector maps to a given output. The equality is exactly the additivity that must be proved.
For instance, with \(f(x)=x+1\) and \(g=f^{-1}\), the incorrect calculation would assert \(g(0+0)=g(0)+g(0)\). Direct evaluation gives \(-1=-2\), which is false. The correction is to identify the assumptions first. When \(T\) is linear and bijective, take the unique preimages \(\mathbf{x}=T^{-1}(\mathbf{u})\) and \(\mathbf{y}=T^{-1}(\mathbf{v})\). Then \(T(\mathbf{x}+\mathbf{y})=\mathbf{u}+\mathbf{v}\), so uniqueness—not an assumed distributive rule for \(T^{-1}\)—proves the required equality.
The same discipline applies to scalar multiplication: establish that \(T(c\mathbf{x})=c\mathbf{u}\), then identify \(c\mathbf{x}\) as the unique preimage. Once the inverse is known to be linear, it can be used as a linear map in its own right. The proof is what licenses that use; the notation \(T^{-1}\) alone does not.
Check Your Understanding
For each question, distinguish what follows from bijectivity and what requires linearity.
- In the proof of additivity, why is \(\mathbf{x}+\mathbf{y}\) a preimage of \(\mathbf{u}+\mathbf{v}\)?
- Which part of bijectivity guarantees that \(T^{-1}(\mathbf{u})\) is defined for every \(\mathbf{u}\in W\), and which part guarantees it is unique?
- For the map \(T(a+bt)=(a+b,a-b)\), compute \(T^{-1}(6,2)\) and verify the result by applying \(T\).
- Give a bijective map from \(\mathbb{R}\) to \(\mathbb{R}\) that is not linear, and explain why the theorem does not apply.
- In one sentence, explain why applying \(T^{-1}\) to a sum does not, by itself, prove that \(T^{-1}\) is additive.
