Theorem. Given that $W$ is finite-dimensional and $T \in \mathcal{L}(V, W)$. then $T'=0$ if and only if $T=0$.
I’m having trouble proving $T=0$ if $T' = 0$. I see a counter example to this. From hypothesis $T'=0$ so equivalently given any linear functional $\psi \in W', T'(\psi)= \psi \circ T=0$. Then $T$ can be non zero when $\psi$ is a zero map i.e. all values in $W$ are mapped to $0$. Then $T$ does not have to be $0$. Could someone please explain what am I missing here? Thank you!
To do that, take any $x\in V$. If $T'=0$ then you have $T'(\phi)=\phi\circ T=0$ for all $\phi\in W'$. Then you have $\phi(Tx)=0$ for all $\phi\in W'$.
So, fix a basis $e_{1},...,e_{m}$ of $W$ and the corresponding dual basis $E_{1},...,E_{m}$ of $W'$. So if $Tx=\sum_{k=1}^{m}c_{k}e_{k}$, then as $E_{k}(Tx)=0$, you have $c_{k}=0$ for each $k$.
This proves that $Tx=0$. Then as $x$ was arbitrary in $V$, you have that $Tx=0$ for all $x$ and hence $T=0$.
PS: This is also true in infinite dimensional normed spaces but it requires something called the Hahn-Banach Theorem to prove it rigorously.