= Solution
If $\phi=T'\psi$ and $v\in\ker T$, then $\phi(v)=\psi(Tv)=0$. Thus $\operatorname{im}T'\subseteq(\ker T)^\circ$, using the <annihilator of a vector subspace>.
Conversely, let $\phi$ vanish on the <kernel of a linear map> $\ker T$. Define a <linear functional> on $\operatorname{im}T$ by $\psi_0(Tv)=\phi(v)$. This is well-defined: $Tv=Tw$ implies $v-w\in\ker T$, so $\phi(v)=\phi(w)$. Extend $\psi_0$ to a <linear functional> $\psi$ on $W$, using the extension argument from part (i). Then $T'\psi=\phi$, proving
$$
\boxed{\operatorname{im}T'=(\ker T)^\circ.}
$$
The dimension formula from part (i) and the <rank-nullity theorem> now give
$$
\operatorname{rank}T'=\dim(\ker T)^\circ
=\dim V-\dim\ker T=\operatorname{rank}T.
$$
To deduce the <eigenvalue> statement, apply this equality of <rank of a linear map> to $T-\lambda I$. Its <dual map> is $T'-\lambda I$, so their <kernel of a linear map> dimensions agree. Thus the same $\lambda$ are <eigenvalues>, with the same <geometric multiplicities>.
For <algebraic multiplicities> as well, part (a)(iii) gives
$$
\bigl((T-\lambda I)^r\bigr)'=(T'-\lambda I)^r,
$$
so equality of <rank of a linear map>[rank] applies to every power. Hence
$$
\dim\ker(T-\lambda I)^r=\dim\ker(T'-\lambda I)^r.
$$
After extending to a <field> where the <characteristic polynomials> split, take $r\geq\dim V$. These <kernel of a linear map> spaces are the <generalized eigenspaces>; the <primary decomposition theorem for linear maps> identifies their dimensions with the <algebraic multiplicities>. The same proof works over the original <field> for every <eigenvalue> in it. We obtain
$$
\boxed{\text{$T$ and $T'$ have the same nonzero eigenvalues,
with both kinds of multiplicity equal.}}
$$
In fact the argument also includes the zero <eigenvalue>.
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