# 6.7: Invertibility

- Page ID
- 277

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**Definition 6.7.1. **A linear map \(T:V\to W \) is called **invertible** if there exists a linear map \(S:W\to V\) such that

\[ TS= I_W \quad \text{and} \quad ST=I_V, \]

where \(I_V:V\to V \) is the identity map on \(V \) and \(I_W:W \to W \) is the identity map on \(W \). We say that \(S \) is an **inverse** of \(T \).

Note that if the linear map \(T \) is invertible, then the inverse is unique. Suppose \(S \) and \(R\) are inverses of \(T \). Then

\begin{equation*}

\begin{split}

ST &= I_V = RT,\\

TS &= I_W = TR.

\end{split}

\end{equation*}

Hence,

\[ S = S(TR) = (ST)R = R. \]

We denote the unique inverse of an invertible linear map \(T \) by \(T^{-1} \).

**Proposition 6.7.2. ***A linear map* \(T\in\mathcal{L}(V,W) \) *is invertible if and only if *\(T \) *is injective and surjective.*

*Proof. *

\(( "\Longrightarrow" )\) Suppose \(T \) is invertible.

To show that \(T \) is injective, suppose that \(u,v\in V \) are such that \(Tu=Tv \). Apply the inverse \(T^{-1} \) of \(T \) to obtain \(T^{-1}Tu=T^{-1}Tv \) so that \(u=v \). Hence \(T \) is injective.

To show that \(T \) is surjective, we need to show that, for every \(w\in W \), there is a \(v\in V\) such that \(Tv=w \). Take \(v=T^{-1}w \in V \). Then \(T(T^{-1}w)=w \). Hence \(T \) is surjective.

\(( "\Longleftarrow" )\) Suppose that \(T \) is injective and surjective. We need to show that \(T \) is invertible. We define a map \(S\in\mathcal{L}(W,V) \) as follows. Since \(T \) is surjective, we know that, for every \(w\in W \), there exists a \(v\in V \) such that \(Tv=w \). Moreover, since \(T \) is injective, this \(v \) is uniquely determined. Hence, define \(Sw=v \).

We claim that \(S \) is the inverse of \(T \). Note that, for all \(w\in W \), we have \(TSw=Tv=w \) so that \(TS=I_W \). Similarly, for all \(v\in V \), we have \(STv=Sw=v \) so that \(ST=I_V \).

Now we specialize to invertible *linear *maps.

**Proposition 6.7.3.** *Let* \(T\in{\cal L}(V,W) \) *be invertible. Then* \(T^{-1}\in {\cal L}(W,V) \).

*Proof.*

Certainly \(T^{-1}:W\longrightarrow V \) so we only need show that \(T^{-1} \) is a linear map. For all \(w_1,w_2\in W \), we have

\[ T(T^{-1}w_1+T^{-1}w_2) = T(T^{-1}w_1) + T(T^{-1}w_2) = w_1 + w_2,\]

and so \(T^{-1}w_1+T^{-1}w_2 \) is the unique vector \(v \) in \(V \) such that \(Tv=w_1+w_2=w \).

Hence,

\begin{equation*}

T^{-1}w_1 + T^{-1}w_2 = v = T^{-1}w = T^{-1}(w_1+w_2).

\end{equation*}

The proof that \(T^{-1}(aw)=aT^{-1}w \) is similar. For \(w\in W \) and \(a\in \mathbb{F} \), we have

\begin{equation*}

T(aT^{-1}w) = a T(T^{-1}w) = aw

\end{equation*}

so that \(aT^{-1}w \) is the unique vector in \(V \) that maps to \(aw \). Hence, \(T^{-1}(aw) = aT^{-1}w \).

**Example 6.7.4. **The linear map \(T(x,y)=(x-2y,3x+y) \) is both injective, since \(\kernel(T) = \{0\} \), and surjective, since \(\range(T) = \mathbb{R}^2 \). Hence, \(T \) is invertible by Proposition 6.7.2.

**Definition 6.7.5.** Two vector spaces \(V \) and \(W \) are called **isomorphic** if there exists an invertible linear map \(T\in\mathcal{L}(V,W) \).

**Theorem 6.7.6. ***Two finite-dimensional vector spaces* \(V \) *and* \(W \) *over* \(\mathbb{F} \) *are isomorphic if and only if* \(\dim(V) = \dim(W) \).

*Proof. *

\(( "\Longrightarrow" )\) Suppose \(V \) and \(W \) are isomorphic. Then there exists an invertible linear map \(T\in\mathcal{L}(V,W) \). Since \(T \) is invertible, it is injective and surjective, and so \(\kernel(T) = \{0\} \) and \(\range(T)= W \). Using the Dimension Formula, this implies that

\[ \dim(V) = \dim(\kernel(T)) + \dim(\range(T)) = \dim(W). \]

\(( "\Longleftarrow" )\) Suppose that \(\dim(V) = \dim(W) \). Let \((v_1,\ldots,v_n) \) be a basis of \(V\) and \((w_1,\ldots,w_n) \) be a basis of \(W \). Define the linear map \(T:V\to W \) as

\[ T(a_1 v_1 + \cdots + a_n v_n) = a_1 w_1 + \cdots + a_n w_n.\]

Since the scalars \(a_1,\ldots,a_n\in\mathbb{F} \) are arbitrary and \((w_1,\ldots,w_n) \) spans \(W \), this means that \(\range(T)=W \) and \(T \) is surjective. Also, since \((w_1,\ldots, w_n) \) is linearly independent, \(T \) is injective (since \(a_1 w_1+\cdots+a_n w_n=0 \) implies that all \(a_1=\cdots=a_n=0 \) and hence only the zero vector is mapped to zero). It follows that \(T \) is both injective and surjective; hence, by Proposition 6.7.2, \(T \) is invertible. Therefore, \(V \) and \(W \) are isomorphic.

We close this chapter by considering the case of linear maps having equal domain and codomain. As in Definition 6.1.1, a linear map \(T \in \mathcal{L}(V,V) \) is called a **linear operator** on \(V \). As the following remarkable theorem shows, the notions of injectivity, surjectivity, and invertibility of a linear operator \(T \) are the same --- as long as \(V \) is finite-dimensional. A similar result does not hold for infinite-dimensional vector spaces. For example, the set of all polynomials \(\mathbb{F}[z] \) is an infinite-dimensional vector space, and we saw that the differentiation map on \(\mathbb{F}[z] \) is surjective, but not injective.

**Theorem 6.7.7. **

*Let*\(V \)

*be a finite-dimensional vector space and*\(T:V\to V \)

*be a linear map. Then the following are equivalent:*

- \(T \)
*is invertible.* - \(T \)
*is injective*. - \(T \)
*is surjective.*

*Proof. *

By Proposition 6.7.2, Part~1 implies Part~2.

Next we show that Part~2 implies Part~3. If \(T \) is injective, then we know that \(\kernel(T)=\{0\} \). Hence, by the Dimension Formula, we have

\[ \dim(\range(T)) = \dim(V) - \dim(\kernel(T)) = \dim(V).\]

Since \(\range(T) \subset V \) is a subspace of \(V \), this implies that \(\range(T) = V \), and so \(T \) is surjective.

Finally, we show that Part~3 implies Part~1. Since \(T \) is surjective by assumption, we have \(\range(T) = V \). Thus, by again using the Dimension Formula,

\[ \dim(\kernel(T)) = \dim(V) - \dim(\range(T)) = 0, \]

and so \(\kernel(T) = \{0\} \), from which \(T \) is injective. By Proposition 6.7.2, an injective and surjective linear map is invertible.

* *

## Contributors

- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis

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