Prove that, $(\text{Im}\ T)^\circ=\text{ker}\ T^\prime$

Question

Prove that, $$(\text{Im}\ T)^\circ=\text{ker}\ T^\prime$$

where,

• $T \in L\ (V,W)$

• $T^\prime \in L\ (W^*,V^*)$

• $\left \langle \varphi \circ T, v \right \rangle = \left \langle T^\prime(\varphi), v \right \rangle = \left [ T^\prime (\varphi)\right ](v) = \varphi\left ( T(v)\right )$, for $\forall \varphi \in W^*$ and $ v \in V$

• $( \text{Im}\ T)^\circ= \big\{\psi \in W^* | \forall T(v) \in \text{Im} \ T, \psi (T(v)) =0 \big\} $

• $\text{ker}\ T^\prime = \big\{\phi \in W^*| T^\prime (\phi)=0\big\}$

Answer 1

We denote,

• $T \in L\ (V,W)$

• $T^\prime \in L\ (W^*,V^*)$

• $\left \langle \varphi \circ T, v \right \rangle = \left \langle T^\prime(\varphi), v \right \rangle = \left [ T^\prime (\varphi)\right ](v) = \varphi\left ( T(v)\right )$, for $\forall \varphi \in W^*$ and $ v \in V$

• $( \text{Im}\ T)^\circ= \big\{\psi \in W^* | \forall T(v) \in \text{Im} \ T, \psi (T(v)) =0 \big\} $

• $\text{ker}\ T^\prime = \big\{\phi \in W^*| T^\prime (\phi)=0\big\}$

because $\psi \in ( \text{Im}\ T)^\circ$ then, $$\psi \in W^*$$ and $$T^\prime(\psi ) = \psi ^\prime$$

for some $\psi ^\prime \in \text{Im} \ T^\prime \in V^*$.

From above we also know that $$\psi^\prime (v)= \left [T^\prime (\psi) \right](v) = \psi \left ( T(v) \right )$$

Because $T(v) \in \text{Im}\ T$

then,

$$\psi \left ( T(v) \right )=0$$

by definition, which means that,

$$ 0 = \psi^\prime (v)$$

for any $v \in V$

Then, $\psi^\prime$ is a zero map which can be expressed as

$$T^\prime ( \psi)=0$$

Then obviously, $\psi \in \text{ker}\ T^\prime$

which means that $$ (\text{Im}\ T)^\circ \subseteq \text{ker}\ T^\prime$$

---

Now, suppose $\phi \in \text{ker}\ T^\prime$ then;

$$\left [T^\prime (\phi)\right ](v)=0(v)=0=\phi (T(v))$$

which means that $$\phi \in (\text{Im}\ T)^\circ$$

since $\phi \in W^*$ and $\phi (T(v))=0$ which satisfies the definition; $$( \text{Im}\ T)^\circ= \big\{\psi \in W^* | \forall T(v) \in \text{Im} \ T, \psi (T(v)) =0 \big\} $$

WE can now deduce that,

$$\text{ker}\ T^\prime \subseteq (\text{Im}\ T)^\circ$$

We have now proved the inclusion in both directions so,

$$(\text{Im}\ T)^\circ=\text{ker}\ T^\prime$$

Answer 2

$(\operatorname{Im} T)^o$ and $\operatorname{Ker} T'$ are both subspaces of $W^*$. For every $f \in W^*$ we have

$$f \in (\operatorname{Im} T)^o \iff f(Tx) = 0, \forall x \in V \iff f \circ T = 0 \iff T'(f) = 0 \iff f \in \operatorname{Ker} T'$$

Therefore $(\operatorname{Im} T)^o = \operatorname{Ker} T'$.