# The Null Space and Range of the Adjoint of a Linear Map

Recall from The Adjoint of a Linear Map page that if $V$ and $W$ are finite-dimensional non-zero vector spaces and if $T \in \mathcal L (V, W)$ then the adjoint of $T$ denoted $T^*$ is defined by considering the linear functional $\varphi: V \to \mathbb{F}$ defined by $\varphi (v) = <T(v), w>$ for a fixed $w \in W$ and we define $T^*(w)$ to be the unique vector in $V$ such that $<T(v), w> = <v, T^*(w)>$.

Also recall from the Orthogonal Complements page that if $U$ is a subset of the vector space $V$ then the orthogonal component of $U$ is the subspace $U^{\perp}$ that is defined to be the set of vectors $v \in V$ such that $v$ is orthogonal to every vector $u \in U$, that is $U^{\perp} = \{ v \in V : <u, v> = 0 , \forall u \in U \}$.

We will now look at the null space and range for adjoint linear maps in the following proposition.

Proposition 1: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $T \in \mathcal L (V, W)$. Then $\mathrm{null} T^* = (\mathrm{range} T)^{\perp}$. |

**Proof:**We must show that $\mathrm{null} T^* \subseteq (\mathrm{range} T)^{\perp}$ and that $\mathrm{null} T^* \supseteq (\mathrm{range} T)^{\perp}$.

- Let $w \in \mathrm{null} T^*$. Then $T^*(w) = 0$. Then for all $v \in V$ we have that $0 = <v, 0> = <v, T^*(w)>$ which implies that $<T(v), w> = 0$ for all $v \in V$. Therefore $w$ is orthogonal to every vector in $\mathrm{range} T$ so $w \in (\mathrm{range} T)^{\perp}$.

- Now let $w \in (\mathrm{range} T)^{\perp}$. Then every vector in $\mathrm{range} T$ is orthogonal to $w$, that is $<T(v), w> = 0$ for all $v \in V$. Thus $<v, T^*(w)> = 0$ for all $v \in V$. But if $<v, T^*(w)> = 0$ for all $v \in V$ then this must imply that $T^*(w) = 0$ so $w \in \mathrm{null} T^*$. $\blacksquare$

Proposition 2:Let $V$ and $W$ be finite-dimensional nonzero inner product spaces. Let $T \in \mathcal L (V, W)$. Then $\mathrm{range} T^* = (\mathrm{null} T)^{\perp}$. |