# Invertibility of a Linear Map Examples 2

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 more examples regarding the invertibility of a linear map.

## Example 1

**Let $T \in \mathcal L (U, V)$ and let $S \in \mathcal L (V, W)$. Let $T$ and $S$ both be invertible. Show that then $ST \in \mathcal L(U, W)$ is invertible with $(ST)^{-1} = T^{-1}S^{-1}$.**

If $T$ is invertible then $T^{-1} \in \mathcal L (V, U)$ exists and if $S$ is invertible then $S^{-1} \in \mathcal (W, V)$ exists.

Now note that:

(1)We also have that:

(2)So $ST$ is invertible and $(ST)^{-1} = T^{-1}S^{-1}$.

## Example 2

**Let $V$ be a finite-dimensional vector space with $\mathrm{dim} (V) ≥ 2$. Show that $\{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$ is not a subspace of $\mathcal L(V, V)$.**

We note that the the zero operator $0(v) = 0$ is not invertible since there exists no linear map $S \in \mathcal L (V, V)$ such that $S0 = I_V$ and $0S = I_V$, so $0 \in \{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$.

We need to show that then either $\{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$ is not closed under addition or that $\{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$ is not closed under scalar multiplication.

Let $\{ v_1, v_2, ..., v_n \}$ be a basis of $V$ with $\mathrm{dim} V = n ≥ 2$. Then for every vector $v \in V$ we have that $v = a_1v_1 + a_2v_2 + ... + a_nv_n$. Define $S \in \mathcal L (V, V)$ by:

(3)Furthermore, define $T \in \mathcal L (V, V)$ by :

(4)Note that $S \in \{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$ since $S$ is not injective. To see this, consider the vector $a_1v_1 + a_2v_2$ and the vector $a_1v_1 + b_2v_2$ where $a_2 \neq b_2$. Then $S(a_1v_1 + a_2v_2) = a_1v_1 = S(a_1v_1 + b_2v_2)$ however $a_1v_1 + a_2v_2 \neq a_1v_1 + b_2v_2$.

We also note that $T \in \{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$ since $T$ is not injective. Once again, to see this, consider the vector $a_1v_1 + a_2v_2$ and the vector $b_1v_1 + a_2v_2$ where $a_1 \neq b_1$. Then $T(a_1v_1 + a_2v_2) = a_2v_2 = T(b_1v_1 + v_2v_2)$ however $a_1v_1 + a_2v_2 \neq b_1v_1 + a_2v_2$.

Notice though that for every vector $v \in V$ we have that:

(5)Therefore $S + T = I_V$. However, the identity map is invertible. Therefore $(S + T) \not \in \{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$.

Thus $\{ T \in \mathcal L(V, V) : T \: \mathrm{is \: not \: invertible} \}$ is not closed under addition and is hence not a subspace of $\mathcal L (V, V)$.