Let $T:\mathbb{R}^{n\times n}\to \mathbb{R}$ be defined as $T(A)=\mathrm{trace}(A)$. Find a basis for $\mathrm{Im}(T)$.
On the one hand, $\mathrm{Im}(A)=t$ so $\mathrm{span}(t)=1$ and therefore the basis is $\{1\}$. On the other hand, $\mathrm{Im}(A)=\mathrm{col}(T)$ but I can not see why $\dim(\mathrm{col}(T))=\mathrm{rank}(T)=1$.
Why can't you see that? The matrix of $\;T\;$ wrt the standard basis in $\;\Bbb R^{n\times n}\,,\,\,\Bbb R\;$ is
$$[T]=\left(1\;0\;0\;\ldots\;0\;1\;0\;\ldots\;1\right)\in R^{n^2}$$
with $\;1\,$-s in the places $\;1\,,\,n+2\,,\,2n+3\,,\,\;$ and etc. Clearly, $\;\dim(Col(T))=1\;$ ...