$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 6.7 Invertibility

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

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:

1. $$T$$ is invertible.
2. $$T$$ is injective.
3. $$T$$ 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

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.