Invertibility of a Linear Map

# Invertibility of a Linear Map

Definition: If $T \in \mathcal L (V, W)$ then the linear map $T$ is said to be Invertible if $\exists S \in \mathcal L (W, V)$ such that $ST = I_V$ and $TS = I_W$. The linear map $S$ is said to be the Inverse Linear Map of $T$ which we denote by $S = T^{-1}$. |

Before we look more into the invertibility of linear maps, we will first look at an important theorem which tells us that if $T \in \mathcal L (V, W)$ is invertible, then the inverse $S = T^{-1}$ is unique.

Theorem 1 (Uniqueness of Inverses): If $T \in \mathcal L (V, W)$ is an invertible linear map, then the inverse linear map $T^{-1} \in \mathcal L (W, V)$ is unique. |

**Proof:**Let $T \in \mathcal L (V, W)$ be an invertible linear map, and suppose that $S, S' \in \mathcal L (W, V)$ are inverses to $T$. Therefore we have that $ST = I_V = S'T$ and $TS = I_W = TS'$.

- So $S = S I_W = S(TS') = (ST)S' = I_V S' = S'$, which implies that $S = S'$ and so the inverse linear map of $T$ is unique. $\blacksquare$

Theorem 2: If $T \in \mathcal L (V, W)$, then $T$ is an invertible linear map if and only if $T$ is bijective. |

**Proof:**$\Rightarrow$ Suppose that $T \in \mathcal L (V, W)$ is an invertible linear map. To show that $T$ is bijective we must show that $T$ is both injective and surjective.

- We will first show that $T$ is injective. Let $u, v \in V$ and suppose that $T(u) = T(v)$. Then we have that $u = T^{-1}(T(u)) = T^{-1}(T(v)) = v$ which implies that $u = v$ and so $T$ is injective.

- Now we will show that $T$ is surjective. Let $w \in W$. So we have that $w = T(T^{-1}(w))$, and $T^{-1} (w) \in V$. So $\forall w \in W$ $\exists T^{-1} (w) \in V$ such that $w = T(T^{-1}(w))$. Therefore $T$ is surjective. We thus conclude that since $T$ is both injective and surjective then $T$ is bijective.

- $\Leftarrow$ Suppose that $T \in \mathcal L (V, W)$ is a bijective linear map. We want to show that $T$ is invertible, which we will show by manually constructing an inverse.

- Let $w \in W$. Since $T$ is surjective, then $\exists v \in V$ such that $w = T(v)$. Now let $S$ be a map, and let $v = S(w)$ be the unique element in $V$ that is mapped to $w$. We note that $v = S(w)$ uniquely maps to $w$ since $T$ is injective. Therefore $w = T(v) = T(S(w))$. Therefore $TS = I_W$.

- We now need to show that $ST = I_V$. Let $v \in V$, and so:

\begin{equation} T(S(T(v)) = T((ST)(v)) = (TS)(T(v)) = I_W T(v) = T(v) \end{equation}

- Since $T$ is injective, we have that $S(T(v)) = v$ and so $ST = I_V$.

- Now we only need to show that $S$ is a linear map, that is show that $S \in \mathcal L (W, V)$. Let $w, x \in W$. Notice that:

\begin{equation} T(S(w) + S(x)) = T(S(w)) + T(S(x)) = w + x \end{equation}

- So $S(w) + S(x)$ is the unique element from $V$ that is mapped to $w + x$. So therefore $S(w) + S(x) = S(w + x)$.

- Lastly, let $a \in \mathbb{F}$. Then $T(aS(w)) = aT(S(w)) = aw$, and so $aS(w)$ is the unique element from $V$ that is mapped to $aw$, so therefore $aS(w) = S(aw)$. Therefore $S$ is a linear map from $W$ to $V$ and so $S \in \mathcal L (W, V)$, so $T$ has an inverse. $\blacksquare$