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

Question

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

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 $\forall \varphi \in W^*$

• $(\text{ker}\ T)^\circ = \big\{\psi \in V^* | \forall u \in \text{ker}\ T, \psi (u)=0\big\}$

• $\text{Im}\ T^\prime = \big\{T^\prime(\varphi)| \varphi \in W^*\big\}$

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 $\forall \varphi \in W^*$

• $(\text{ker}\ T)^\circ = \big\{\psi \in V^* | \forall u \in \text{ker}\ T, \psi (u)=0\big\}$

• $\text{Im}\ T^\prime = \big\{T^\prime(\varphi)| \varphi \in W^*\big\}$

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Suppose that $\phi \in \text{Im}\ T^\prime$ such that $\phi = T^\prime(\varphi) $, then; $$\phi (v)= \left [ T^\prime(\varphi)\right ](v) = \varphi\left ( T(v)\right )$$

Because $u \in V$ as $\text{ker}\ T \subseteq V$

$$\phi (u)=\left [ T^\prime(\varphi)\right ](u) = \varphi\left ( T(u)\right )$$

But $T(u) =0$ by definition, so $$\phi (u)=\left [ T^\prime(\phi)\right ](u ) = \varphi\left ( 0\right )=0$$ .

Because

$$(\text{ker}\ T)^\circ = \big\{\psi \in V^* | \forall u \in \text{ker}\ T, \psi (u)=0\big\}$$

and

$$\phi \in\text{Im}\ T^\prime \in V^*$$

where $\phi (u) =0$

so by definition, $\phi \in (\text{ker}\ T)^\circ$

Also, $\phi $ can be any element of $\text{Im}\ T^\prime$, the above relation implies that $\text{Im}\ T^\prime \subseteq (\text{ker}\ T)^\circ$

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Now, we know that $\text{dim Im}\ T^\prime= \text{dim Im}\ T$

then, $$\text{dim Im}\ T^\prime= \text{dim}\ V-\text{dim ker}\ T$$ because $$\text{dim Im}\ T = \text{dim}\ V-\text{dim ker}\ T $$ from the rank nullity theorem.

We also know that, $\text{dim}\ V= \text{dim ker}\ T + \text{dim} (\text{ker}\ T)^\circ $.

$\therefore \text{dim Im}\ T^\prime= (\text{dim ker}\ T + \text{dim} (\text{ker}\ T)^\circ)-\text{dim ker}\ T = \text{dim} (\text{ker}\ T)^\circ)$

We can now conclude that $$\text{Im} \ T^\prime= ( \text{ker}\ T)^\circ$$

$\text{QED}$