[proofplan] By the [First Isomorphism Theorem for Vector Spaces](/theorems/384), the quotient $U/\ker\alpha$ is isomorphic to $\mathrm{im}\,\alpha$. Since isomorphic finite-dimensional spaces have the same dimension, and the quotient dimension formula gives $\dim(U/\ker\alpha) = \dim U - \dim\ker\alpha$, we obtain $r(\alpha) + n(\alpha) = \dim U$ by combining these two equalities. [/proofplan] [step:Apply the First Isomorphism Theorem to obtain $U/\ker\alpha \cong \mathrm{im}\,\alpha$] By the [First Isomorphism Theorem for Vector Spaces](/theorems/384), the map $\vartheta: U/\ker\alpha \to \mathrm{im}\,\alpha$ defined by $\vartheta(u + \ker\alpha) = \alpha(u)$ is an isomorphism of $\mathbb{F}$-vector spaces. [/step] [step:Equate dimensions using the quotient formula and the isomorphism] Since $U$ is finite-dimensional, $\ker\alpha$ is a subspace of $U$ and hence finite-dimensional by [Dimension of Subspaces](/theorems/375). By the [Rank-Nullity for Quotient Spaces](/theorems/377): \begin{align*} \dim(U/\ker\alpha) = \dim U - \dim\ker\alpha. \end{align*} Since $\vartheta$ is an isomorphism, $\mathrm{im}\,\alpha$ is finite-dimensional with $\dim\mathrm{im}\,\alpha = \dim(U/\ker\alpha)$ by the [Finite Dimensional Isomorphism Criterion](/theorems/380). [guided] We now connect the three dimension equalities. The isomorphism $\vartheta: U/\ker\alpha \xrightarrow{\sim} \mathrm{im}\,\alpha$ tells us that $U/\ker\alpha$ and $\mathrm{im}\,\alpha$ are isomorphic. By the [Finite Dimensional Isomorphism Criterion](/theorems/380), isomorphic vector spaces have the same dimension, so \begin{align*} \dim\mathrm{im}\,\alpha = \dim(U/\ker\alpha). \end{align*} Meanwhile, $\ker\alpha$ is a subspace of the finite-dimensional space $U$, so it is finite-dimensional by [Dimension of Subspaces](/theorems/375). The [Rank-Nullity for Quotient Spaces](/theorems/377) then gives \begin{align*} \dim(U/\ker\alpha) = \dim U - \dim\ker\alpha. \end{align*} Combining: $\dim\mathrm{im}\,\alpha = \dim U - \dim\ker\alpha$. [/guided] [/step] [step:Conclude $r(\alpha) + n(\alpha) = \dim U$] Substituting the definitions $r(\alpha) = \dim\mathrm{im}\,\alpha$ and $n(\alpha) = \dim\ker\alpha$: \begin{align*} r(\alpha) = \dim U - n(\alpha), \end{align*} which rearranges to $r(\alpha) + n(\alpha) = \dim U$. [/step]