[proofplan] The equivalence (1) $\Leftrightarrow$ (2) is definitional. For (2) $\Leftrightarrow$ (3), any eigenvector basis partitions into eigenspaces, and [Linear Independence of Eigenvectors](/theorems/403) ensures the sum is direct. The equivalence (3) $\Leftrightarrow$ (4) is a dimension count using the directness of the eigenspace sum. [/proofplan] [step:Show (1) $\Leftrightarrow$ (2) by definition of diagonalisability] By definition, $\alpha$ is diagonalisable if and only if there exists a basis of $V$ in which the matrix of $\alpha$ is diagonal. A diagonal matrix has the eigenvalues $\lambda_1, \dots, \lambda_n$ on the diagonal, and the corresponding basis vectors are eigenvectors. Conversely, if $V$ has a basis of eigenvectors, the matrix of $\alpha$ in that basis is diagonal. [/step] [step:Show (2) $\Rightarrow$ (3) using directness of eigenspace sums] Suppose $V$ has a basis of eigenvectors $(b_1, \dots, b_n)$. Each $b_j$ lies in some eigenspace $E_\alpha(\lambda_i)$, so $V = E_\alpha(\lambda_1) + \cdots + E_\alpha(\lambda_k)$. By [Linear Independence of Eigenvectors](/theorems/403), this sum is direct: \begin{align*} V = E_\alpha(\lambda_1) \oplus \cdots \oplus E_\alpha(\lambda_k). \end{align*} [/step] [step:Show (3) $\Rightarrow$ (2) by assembling bases of eigenspaces] If $V = \bigoplus_{i=1}^k E_\alpha(\lambda_i)$, choose a basis $\mathcal{B}_i$ for each $E_\alpha(\lambda_i)$. The union $\mathcal{B}_1 \cup \cdots \cup \mathcal{B}_k$ is a basis for $V$ by the definition of direct sum. Every vector in this basis is an eigenvector, establishing (2). [/step] [step:Show (3) $\Leftrightarrow$ (4) by a dimension count] By [Linear Independence of Eigenvectors](/theorems/403), the sum $E_\alpha(\lambda_1) + \cdots + E_\alpha(\lambda_k)$ is always direct. Hence the sum equals $V$ if and only if the dimensions add up: \begin{align*} \dim V = \dim(E_\alpha(\lambda_1) \oplus \cdots \oplus E_\alpha(\lambda_k)) = \sum_{i=1}^k \dim E_\alpha(\lambda_i). \end{align*} This is exactly condition (4). [/step]