6.4 Invertibility
Definition 6.4.1. A \(T\in \mathcal {L}(A,B)\) is called invertible if there exists \(S\in \mathcal {L}(B,A)\) such that \(ST=I\) on \(A\) and \(TS=I\) on \(B\). Then \(S\) is called the inverse of \(T\) and we write \(S=T^{-1}\); here \(I\) denotes the identity map.
The inverse is unique. To see this, let \(S'\) be another inverse of \(T\); then \[S'=S'I=S'(TS)=(S'T)S=IS=S.\]
Proposition 6.4.2. A linear transformation is invertible if and only if it is both injective and surjective.
Proof. Let \(T\in \mathcal {L}(A,B)\).
Invertible \(\implies \) injective and surjective
Suppose \(S=T^{-1}\) exists. If \(Tx=Ty\) then applying \(S\) gives \(x=(ST)x=(ST)y=y\), so \(T\) is injective. And for any \(b\in B\), the vector \(x=Sb\) satisfies \(Tx=(TS)b=b\), so \(T\) is surjective.
Injective and surjective \(\implies \) invertible
Suppose \(T\) is bijective. As a map of sets it has an inverse \(S\) — Theorem 1.3.8 — defined by letting \(S(b)\) be the unique \(a\in A\) with \(Ta=b\). What has to be checked is that this \(S\) is linear, since a linear transformation is required to have a linear inverse.
Let \(b_1,b_2\in B\) and put \(a_1=Sb_1\), \(a_2=Sb_2\), so \(Ta_1=b_1\) and \(Ta_2=b_2\). By linearity of \(T\), \[T(a_1+a_2)=Ta_1+Ta_2=b_1+b_2,\] and since \(S\) returns the unique preimage, \[S(b_1+b_2)=a_1+a_2=Sb_1+Sb_2.\] Similarly \(T(\lambda a_1)=\lambda Ta_1=\lambda b_1\) gives \(S(\lambda b_1)=\lambda a_1=\lambda Sb_1\). So \(S\) is linear, and \(ST=I\), \(TS=I\) by construction. Hence \(T\) is invertible. □
Remark. The content here is entirely in the second half, and specifically in the linearity of \(S\): bijectivity alone gives an inverse function, and it is a small theorem — not a definition — that this function respects the vector space operations.
Definition 6.4.3. Two vector spaces are called isomorphic if there is an invertible linear transformation from one vector space onto the other.
Remark.
- 1.
- Two isomorphic spaces have the same properties.
- 2.
- The Greek word Iso means equal, morph means shape. Thus isomorphic laterally means equal shape.
- 3.
- Two finite dimensional vector spaces are isomorphic if and only if they have the same dimension.
Definition 6.4.4. An operator is \(LT\) from a vector space to itself i.e. \(A\) \(LT,\) \(T:A\longrightarrow A\).
\(\mathcal {L}(A)\) is the set of all operators on a vector space \(A\).
Note that \(\mathcal {L}(A)=\mathcal {L}(A,A)\). \(T\in \mathcal {L}(A)\) means \(T\) is an operator on a vector space \(A\).
Theorem 6.4.5. If \(A\) is finite dimensional and if \(T\in \mathcal {L}(A)\). Then the following are equivalent.
- 1.
- \(T\) is invertible
- 2.
- \(T\) is injective
- 3.
- \(T\) is surjective
Proof. Let \(T\in \mathcal {L}(A)\) where \(A\) is finite dimensional.
\((1)\implies (2)\) If \(T\) is invertible then it injective by Proposition 6.4.2.
\((2)\implies (3)\) If \(T\) is injective then null\(T=\{0\}\) by Proposition 6.2.4.
From Theorem 6.2.8, \(\dim \) range\(T=\dim A-\dim \) null\(T=\dim A-0=\dim A\).
\(\dim \) range\(T=\dim A\implies \) range\(T=A\)
\(\implies T\) is surjective.
\((3)\implies (1)\) If \(T\) is surjective then range\(T=A\), thus \(\dim \) null\(T=\dim A-\dim \) range\(T=0\).
\(\implies \) null\(T=\{0\}\).
\(\implies T\) is injective.
Since now \(T\) is both injective and surjective \(\implies T\) is invertible.
□
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