[proofplan] All three equivalences follow from the [Rank-Nullity Theorem](/theorems/385) applied with $\dim U = \dim V$. Injectivity means $n(\alpha) = 0$, which forces $r(\alpha) = \dim V$, i.e., surjectivity. Conversely, surjectivity forces $n(\alpha) = 0$, i.e., injectivity. The equivalence with isomorphism then follows from [Isomorphism iff Linear Bijection](/theorems/378). [/proofplan] [step:Prove injectivity implies surjectivity via rank-nullity] Suppose $\alpha$ is injective. Then $\ker\alpha = \{\mathbf{0}\}$, so $n(\alpha) = 0$. By the [Rank-Nullity Theorem](/theorems/385): \begin{align*} r(\alpha) = \dim U - n(\alpha) = \dim U = \dim V. \end{align*} Since $\mathrm{im}\,\alpha \subseteq V$ is a subspace with $\dim\mathrm{im}\,\alpha = \dim V$, we have $\mathrm{im}\,\alpha = V$ by [Dimension of Subspaces](/theorems/375). Hence $\alpha$ is surjective. [guided] If $\alpha$ is injective, its kernel is trivial: $\ker\alpha = \{\mathbf{0}\}$, giving $n(\alpha) = 0$. The [Rank-Nullity Theorem](/theorems/385) states $r(\alpha) + n(\alpha) = \dim U$, so \begin{align*} r(\alpha) = \dim U - 0 = \dim U = \dim V, \end{align*} where the last equality uses the hypothesis $\dim U = \dim V$. Now $\mathrm{im}\,\alpha$ is a subspace of $V$ with dimension equal to $\dim V$. By [Dimension of Subspaces](/theorems/375), a subspace of a finite-dimensional space whose dimension equals that of the ambient space must be the entire space. Therefore $\mathrm{im}\,\alpha = V$, and $\alpha$ is surjective. The key mechanism: in finite dimensions with equal domain and codomain dimensions, the rank and nullity are complementary -- increasing one forces the other to decrease. Zero nullity leaves the rank no choice but to be maximal. [/guided] [/step] [step:Prove surjectivity implies isomorphism] Suppose $\alpha$ is surjective, so $\mathrm{im}\,\alpha = V$ and $r(\alpha) = \dim V = \dim U$. By the [Rank-Nullity Theorem](/theorems/385): \begin{align*} n(\alpha) = \dim U - r(\alpha) = \dim U - \dim U = 0. \end{align*} Hence $\ker\alpha = \{\mathbf{0}\}$, so $\alpha$ is injective. Since $\alpha$ is both injective and surjective, it is bijective, and hence an isomorphism by the [Isomorphism iff Linear Bijection](/theorems/378) theorem. [/step] [step:Prove isomorphism implies injectivity] If $\alpha$ is an isomorphism, then by definition it is a bijective [linear map](/page/Linear%20Map), hence in particular injective. [/step]