Invertibility of a Linear Map Examples 1

# Invertibility of a Linear Map Examples 1

Let $V$ and $W$ be vector spaces. Recall from the Invertibility of a Linear Map page that a linear map $T \in \mathcal L (V, W)$ is said to be invertible if there exists a linear map $S \in \mathcal L (W, V)$ such that $ST = I_V$ and $TS = I_W$ where $I_V$ is the identity map on $V$ and $I_W$ is the identity map on $W$. Furthermore, the linear map $S = T^{-1}$ is said to be the inverse of $T$.

We will now look at some examples regarding the invertibility of a linear map.

## Example 1

Reprove that if $T \in \mathcal L (V, W)$ is invertible then $T$ has a unique inverse $S \in \mathcal L (W, V)$.

Suppose that $S, S' \in \mathcal L (W, V)$ are both inverses to $T$ - that is $ST = I_V$ and $TS = I_W$ as well as $S'T = I_V$ and $TS' = I_W$. Then we have that:

(1)
\begin{align} \quad S = SI_W = S(TS') = (ST)S' = I_VS' = S' \end{align}

Therefore $S = S'$ so the inverse $S$ is unique.

## Example 2

Reprove that if $T \in \mathcal L(V, W)$ then $T$ is invertible if and only if $T$ is bijective.

$\Rightarrow$ Suppose that $T$ is invertible. Then $T^{-1} \in \mathcal L (W, V)$ exists such that $T^{-1} T = I_V$ and $TT^{-1} = I_W$. We need to show that then $T$ is both injective and surjective.

Suppose that $T(u) = T(v)$. Then we have that:

(2)
\begin{align} \quad u = T^{-1}T(u) = T^{-1}T(v) = v \end{align}

Therefore $u = v$ so $T$ is injective.

To show that $T$ is surjective, note that for every vector $w \in W$ we have that $w = TT^{-1}(w) = T(T^{-1}(w))$. Therefore, for every vector $w \in W$ there exists a vector $T^{-1}(w) \in V$ such that $w = T(T^{-1}(w))$, so $T$ is surjective.

$\Leftarrow$ Now suppose that $T$ is bijective. Then $T$ is both injective and surjective. We will show that $T$ is invertible by constructing a linear map $S$.

Define $S : W \to V$ to be $v = S(w)$ where $v$ is the unique vector $v \in V$ that is mapped to $w \in W$. We note that for every vector $w \in W$ there exists a vector $v \in V$ such that $w = T(v)$ since $T$ is surjective, and the vector $v$ is unique for every vector $w$ since $T$ is surjective.

Then clearly:

(3)
\begin{align} \quad T(v) = T(S(w)) = (TS)(w) = w \end{align}

Therefore $(TS)(w) = w$ implies that $TS = I_W$. Furthermore we note that:

(4)
$$T(S(T(v)) = (TS)T(v) = I_wT(v) = T(v)$$

The equation above says that $T(S(T(v))) = T(v)$, and since $T$ is injective this implies that $S(T(v)) = v$ so $(ST)(v) = v$ and therefore $ST = I_V$.

We will not reprove that $S$ is a linear map since that follows easily from showing that the additivity and homogeneity properties of linear maps hold.

## Example 3

Prove that the linear map $T \in \mathcal L (\wp (\mathbb{R}), \wp (\mathbb{R}))$ defined by $T(p(x)) = x^2 p(x)$ for every $p(x) \in \wp (\mathbb{R})$ is not invertible.

Consider the polynomial $1 \in \wp (\mathbb{R})$. Note that there exists no polynomial $p(x) \in \wp (\mathbb{R})$ such that $T(p(x)) = x^2 p(x) = 1$. Therefore $T$ is not surjective which implies that $T$ is not invertible.