- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.6 Inverses
$\quad$ Given a function $T,$ we want to find (if possible) another function $S$ whose composition with $T$ is the identity transformation. Since composition is not commutative in general, we must distinguish between $ST$ and $TS.$ Therefore, we introduce two kinds of inverses - left and right inverses.
$\quad$Definition.$\quad$ Given two sets $V$ and $W$ and a function $T: V \to W,$ a function $S: T(V) \to V$ is called a left inverse of $T$ if $S[T(x)] = x$ for all $x$ in $V.$ That is, if $$ST = I_V,$$ where $I_V$ is the identity transformation on $V.$ A function $R: T(V) \to V$ is called a right inverse of $T$ if $T[R(y)] = y$ for all $y$ in $T(V).$ That is, if $$TR = I_{T(V)},$$ where $I_{T(V)}$ is the identity transformation on $T(V).$
$\quad$ Every function $T:V \to W$ has at least one right inverse. In fact, each $y$ in $T(V)$ has the form $y = T(x)$ for at least one $x$ in $V.$ If we select one such $x$ and define $R$ so that $R(y) = x,$ then $T[R(y)] = T(x) = y$ for each $y$ in $T(V),$ so $R$ is a right inverse. Nonuniqueness may occur because there may be more than one $x$ in $V$ which maps onto a given $y$ in $T(V).$
$\quad$ We will prove in Theorem 2.9 that if each $y$ in $T(V)$ is the image of exactly one $x$ in $V,$ then the right inverses are unique. First we will prove that if a left inverse exists - it is unique, and at the same time - it is a right inverse.
$\quad$Theorem 2.8.$\quad$ A function $T:V \to W$ can have at most one left inverse. If $T$ has a left inverse $S,$ then $S$ is also a right inverse.
$\quad$Proof.$\quad$ Assume $T$ has two left inverses, $S: T(V) \to V$ and $S': T(V) \to V.$ Now, choose any $y$ in $T(V).$ We will prove that $S(y) = S'(y).$ For some $x$ in $V,$ we have $T(x) = y.$ Thus, $S(y) = S[T(x)]$ and $S'(y) = S'[T(x)].$ But because $S$ and $S'$ are left inverses, we know by definition that \begin{align*} S[T(x)] &= x \qquad \text{and} \qquad S'[T(x)] = x \end{align*} Thus, we have $S(y) = S'(y).$ But since this is the case for any $y$ in $T(V),$ it follows that $S = S',$ and the left inverse of $T$ is unique.
$\quad$ We will now prove that every left inverse is also a right inverse. Recall that $S: T(V) \to V.$ Let $y$ be any element in $T(V).$ For some $x$ in $V,$ we have $y = T(x),$ and because $S$ is the left inverse of $T,$ we have $$S(y) = ST(x) = x.$$ If we take the composition of $T$ with the above equation, we find that \begin{align*} T[S(y)] &= T[ST(x)] \\ &= T(x) \end{align*} But as we saw before, $T(x) = y,$ making $TS(y) = y,$ thus proving that $S$ is a right-inverse of $T. \quad \blacksquare$
$\quad$Theorem 2.9.$\quad$ A function $T:V \to W$ has a left inverse if and only if $T$ maps distinct elements of $V$ onto distinct elements of $W;$ that is, if and only if, for all $x$ and $y$ in $V,$ \begin{align*} (2.5) \qquad x &\neq y \qquad \text{implies} \qquad T(x) \neq T(y) \end{align*} $\qquad$Note:$\quad$ Condition (2.5) is equivalent to the statement \begin{align*} (2.6) \qquad T(x) &= T(y) \qquad \text{implies} \qquad x = y \end{align*} $\quad$ A function satisfying (2.5) or (2.6) is said to be one-to-one on $V.$
$\quad$Proof.$\quad$ First, assume that $T: V \to W$ has a left inverse $S: T(V) \to V.$ We will show that this implies $T$ is one-to-one on $V.$ Let $x$ and $y$ be elements of $V$ such that $T(x) = T(y).$ If we compose $S$ with both sides of this equation, we get \begin{align*} S[T(x)] = S[T(y)] \end{align*} But we know that $S$ is the left inverse of $T,$ which means the above equation is equivalent to $x = y.$ This proves that if $T$ has a left inverse, then $T(x) = T(y)$ implies $x = y.$
$\quad$ Now, we will prove that if $T$ is one-to-one on $V,$ then there exists a function $S: T(V) \to V$ which is a left inverse to $T.$ Suppose $y$ is an element of $T(V).$ Because $T$ is one-to-one on $V,$ there is exactly one $x$ in $V$ such that $T(x) = y.$ If we define $S:T(V) \to V$ as \begin{align*} S(y) &= x \qquad \text{where} \qquad T(x) = y \end{align*} then we have $S(y) = S[T(x)] = x,$ for all $x$ in $V,$ which in turn implies that $ST = I_V.$ Thus, $S$ is a left inverse of $T.$ This completes the proof.
$\quad$Definition.$\quad$ Let $T: V \to W$ be one-to-one on $V.$ The unique left inverse of $T$ (which we know is also a right inverse) is denoted by $T^{-1}.$ We say that $T$ is invertible, and that $T^{-1}$ is the inverse of $T.$
$\quad$ The results of this section refer to arbitrary functions. In the next section, we will apply these ideas to linear transformations.