Let $T : \mathbb{R}^n \xrightarrow{} \mathbb{R}^n$. The domain of $T$ is finite dimensional $\mathbb{R}^n$, so by rank nullity theorem we have $n=$ dim(ker($T$)) + dim(im($T$)).
Although I do not know how to prove it, I found on Wikipedia that ("via the splitting lemma") the domain is isomorphic to the direct sum of the kernel and image, i.e. $\mathbb{R}^n\simeq \text{ker}(T) \oplus \text{im}(T)$. To clarify, when I say the "sum" I mean the set of all vectors expressible as the vector addition of a vector in the kernel and a vector in the image (both are subspaces of $\mathbb{R}^n$ in this case), and "direct sum" means this expression is unique for each vector in the sum.
Since the sum of the kernel and image is a subspace of $\mathbb{R}^n$ with the same dimension as $\mathbb{R^n}$, is the isomorphic relationship actually an equality in this case? I.e. $\text{ker}(T) \oplus \text{im}(T) = \mathbb{R}^n$?
Second, does this set equality hold more generally for linear transformations $T:V \xrightarrow{} V$, where $V$ is a finite dimensional vector space over $\mathbb{R}$? I.e. $\text{ker}(T) \oplus \text{im}(T) = V$?
No. Take, for instance,$$\begin{array}{rccc}T\colon&\mathbb R^2&\longrightarrow&\mathbb R^2\\&(x,y)&\mapsto&(0,x).\end{array}$$Then $\ker T=\operatorname{im}T=\{0\}\times\mathbb R$.