Invertible Transformations

Left and right invertibility

For more on left and right inverses, look up exercise 1 here.

Left inverse

Let $A:V\to W$ be a linear transformation. It is said to be left invertible if there exists a linear transformation $B:W\to V$ such that $BA=I_V$. $A$ must be injective for $B$ to exist.

Right inverse

The transformation $A$ is called right invertible if there exists a linear transformation $C:W\to V$ such that $AC=I_W$. $A$ must be surjective for $C$ to exist.

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Properties and examples

$B$ and $C$ are called the left and right inverses of $A$.
Generally, $B$ and $C$ are not unique.

Examples:

• The LT $\left[\begin{array}{c}1\\ 1\end{array}\right]$ (which maps from $\mathbb{R}\to {\mathbb{R}}^2$) is left invertible, but not right invertible. Notice that it is injective, i.e, only one vector in ${\mathbb{R}}^2$ is associated with every vector in $\mathbb{R}$. However, it is not surjective, since it cannot map to vectors which are not of the form $\left(\begin{array}{c}x\\ x\end{array}\right)$. It has infinitely many left inverses of the form $\left[\begin{array}{cc}a& 1-a\end{array}\right]$.
• The LT $\left[\begin{array}{cc}1& 1\end{array}\right]$ (which maps from ${\mathbb{R}}^2\to \mathbb{R}$) is right invertible, but not left invertible. Note that it is surjective, with every vector in $\mathbb{R}$ being mapped to. It is clearly not injective, with every vector lying on the line $x+y=a,\left(\begin{array}{c}x\\ y\end{array}\right)\in {\mathbb{R}}^2$ being mapped to $a\in \mathbb{R}$. It also has infinitely many right inverses of the form $\left[\begin{array}{c}b\\ 1-b\end{array}\right]$.

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Invertible transformations

Definition

Definition 1(Definition).

A linear transformation $A:V\to W$ is called invertible if it is both right and left invertible.

Coincidence of left and right inverses

Theorem 2(Theorem).

If a linear transformation $A:V\to W$ is invertible, then its left and right inverses $B$ and $C$ are unique and coincide.

Proof
Let $BA=I$ and $AC=I$. Then,

$$BAC=B(AC)=BI=B$$

also

$$BAC=(BA)C=IC=C$$

therefore $B=C$.

Suppose for some transformation $B_1$ we have $B_{1}A=I$. repeating the above reasoning with $B_1$ instead of $B$ we get $B_1=C$. Therefore the left inverse $B$ is unique. A similar reasoning can be applied to $C$. ❏

Corollary

A transformation $A:V\to W$ is invertible if and only if there exists a unique linear transformation (denoted $A^{-1}$), $A^{-1}:W\to V$, such that

$$A^{-1}A=I_V,A{A}^{-1}=I_W$$

It is clear from the above corollary that if $A$ is invertible, $A^{-1}$ is also invertible.

Theorem 3(Theorem).

An invertible matrix must be square. Moreover, if a square matrix has either a left or right inverse, it is invertible.

Proof: TBD

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Inverse of a product/composition of transformations

Theorem 4(Theorem).

If LTs $A$ and $B$ are invertible (and the product $AB$ is defined), then the product $AB$ is invertible, and $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

Proof

$$(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=AI{A}^{-1}=A{A}^{-1}=I,$$

and similarly,

$$(B^{-1}{A}^{-1})AB=B^{-1}(A^{-1}A)B=B^{-1}IB=B^{-1}B=I.$$

❏

The invertibility of the product $AB$ does not imply $A$ and $B$ are invertible. However,

Theorem 5(Theorem).

If one of the factors (either $A$ or $B$) and the product $AB$ is invertible, then the second factor is invertible too.

Proof
Assume $B$ and $AB$ are invertible.

$$\begin{array}{rl}(AB{)}^{-1}(AB)& =I\\ B(AB{)}^{-1}(AB)B^{-1}& =BI{B}^{-1}\\ B(AB{)}^{-1}A& =I\end{array}$$

Thus, $A$ has a left inverse, namely $B(AB{)}^{-1}$. Also,

$$\begin{array}{rl}(AB)(AB{)}^{-1}& =I\\ A(B(AB{)}^{-1})& =I\end{array}$$

Thus, $A$ has a right inverse, namely $B(AB{)}^{-1}$. Therefore, $A$ is invertible. ❏

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Inverse of transpose

Theorem 6(Theorem).

If a matrix $A$ is invertible, then $A^T$ is also invertible and $(A^T{)}^{-1}=(A^{-1}{)}^T$

The proof is trivial.