# Kernels as subspaces of vector space of all linear transformations

Let $V$ be a finite dimensional vector space and let $T : V \rightarrow V$ be a linear transformation. Let $W$ be the vector space of all linear transformations $S : V \rightarrow V$ and let $L_{T}$ and $R_{T}$ be the linear transformations $W \rightarrow W$ defined by $L_{T}(S)=TS$, $R_{T}(S)=ST$. Prove that the kernels of $L_{T}$ and $R_{T}$ have the same dimension.

It is straightforward to show that $\ker L_{T} = \text{Hom}_{K}(V, \ker T)$. So it remains to show that $\text{dim} \ker R_{T} = \text{dim}V \cdot \text{dim} \ker T$. For this, I need a linear isomorphism between $\ker R_{T}$ and $\text{Hom}_{K}(V, \ker T)$, which I am not seeing.

Just dualize your first example: $\ker R_T = \text{Hom}(\text{coker} T,V),$ where $\text{coker}$ means cokernel, $\text{coker} T = V/\text{Im} T.$ Then since $\dim\text{coker} T = \dim V - \dim \text{Im} T = \dim \ker T$ by first isomorphism theorem, you have your result.