Let $T: \mathbb{R}^6 \rightarrow \mathbb{R}^3$ be an onto linear transformation (that is, surjective)
What is the minimal value for $\dim(\ker(T))$ , what is the maximal value for $\dim(\operatorname{im}(T))$?
Because $T$ is onto it needs to span all of $\mathbb{R}^3$; therefore $\dim(\operatorname{im}(T)) \leq 6 $ and $\dim(\ker(T)) \geq 0$ .
Best Answer
How can $\dim \operatorname{Im}(T)$ be greater than the dimension of the codomain?
Note that, if $T$ is surjective (on-to), then $\operatorname{Im}(T)=\mathbb R^3$. From this, you can deduce the dimension of the image and apply the rank-nullity theorem to deduce the dimension of the kernel.