# Injectivity and Surjectivity of the Adjoint of a Linear Map

In the following two propositions we will see the connection between a linear map $T$ being injective/surjective and the corresponding adjoint matrix $T^*$ being surjective/injective.

Proposition 1: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces and let $T \in \mathcal L (V, W)$. Then $T$ is injective if and only if $T^*$ is surjective. |

**Proof:**$\Rightarrow$ Suppose that $T$ is an injective. We first note that $T^* : W \to V$.

- Since $T$ is injective, we have that $\mathrm{null} (T) = \{ 0 \}$. We’ve already seen that $\mathrm{null} (T) = (\mathrm{range} (T^*))^{\perp}$, and so $(\mathrm{range} (T^*))^{\perp} = \{ 0 \}$. If we apply the orthogonal complement to both sides of the equation above, then we get that:

(1)

\begin{align} \quad (\mathrm{range} (T^*))^{\perp} = \{ 0 \} \\ \quad ((\mathrm{range} (T^*))^{\perp})^{\perp} = \{ 0 \}^{\perp} \\ \quad \mathrm{range} (T^*) = V \end{align}

- Therefore $T^*$ is surjective.

- $\Leftarrow$ Now suppose that $T^*$ is surjective. Then $V = \mathrm{range} (T^*)$. We already known that $\mathrm{range} (T^*) = (\mathrm{null} (T))^{\perp}$ and so $V = (\mathrm{null} (T))^{\perp}$. If we apply the orthogonal complement to both sides of the equation above, then we get that:

(2)

\begin{align} \quad V = (\mathrm{null} (T))^{\perp} \\ \quad V^{\perp} = ((\mathrm{null} (T))^{\perp})^{\perp} \\ \quad \{ 0 \} = \mathrm{null} (T) \end{align}

- Therefore $T$ is injective. $\blacksquare$

Proposition 2: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces and let $T \in \mathcal L (V, W)$. Then $T$ is surjective if and only if $T^*$ is injective. |

**Proof:**$\Rightarrow$ Suppose that $T$ is surjective. Then $V = \mathrm{range} (T)$. But we already know that $\mathrm{range} (T) = (\mathrm{null} (T^*))^{\perp}$ and so $V = (\mathrm{null} (T^*))^{\perp}$. If we apply the orthogonal complement to both sides of the equation above, then we get that:

(3)

\begin{align} \quad V = (\mathrm{null} (T^*))^{\perp} \\ \quad V^{\perp} = ((\mathrm{null} (T^*))^{\perp})^{\perp} \\ \quad \{0 \} = \mathrm{null} (T^*) \end{align}

- Therefore $T^*$ is injective.

- $\Leftrightarrow$ Suppose that $T^*$ is injective. Then $\mathrm{null} (T^*) = \{ 0 \}$. However, we know that $\mathrm{null} (T^*) = (\mathrm{range} (T))^{\perp}$ so $(\mathrm{range} (T))^{\perp} = \{ 0 \}$. If we apply the orthogonal complement to both sides of the equation above, then we get that:

(4)

\begin{align} \quad (\mathrm{range} (T))^{\perp} = \{ 0 \} \\ \quad ((\mathrm{range} (T))^{\perp})^{\perp} = \{ 0 \}^{\perp} \\ \quad \mathrm{range} (T) = V \end{align}

- Therefore $T$ is surjective. $\blacksquare$