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  "body": "\n## Proof\n\nWe build a basis of eigenvectors.\n\n1. Let $\\lambda_1, \\dots, \\lambda_n \\in F$ be pairwise distinct [[eigenvalues|0ZND7NQ970]] of $A$, that is, of the map $x \\mapsto A x$ on $F^n$.\n2. For each $i$, choose an [[eigenvector|0B5H4208Z9]] $v_i \\in F^n$ for $\\lambda_i$. So $v_i \\ne 0$ and $A v_i = \\lambda_i v_i$.\n3. The eigenvalues are pairwise distinct. So $v_1, \\dots, v_n$ are linearly independent, because [[eigenvectors with distinct eigenvalues are linearly independent|0HNMFG8F8K]].\n4. The space $F^n$ has [[dimension $n$|0QK7FPE6VE]]. So the $n$ independent vectors $v_1, \\dots, v_n$ form a [[basis|096V7H13H9]] of $F^n$.\n5. Thus $F^n$ has a basis of eigenvectors of $A$. By the [[characterisation of diagonalisable matrices|056SW1E0AC]], $A$ is [[diagonalisable|0Y17PEKF24]] over $F$.\n\n## Idea\n\nOne eigenvector for each of the $n$ distinct eigenvalues already gives $n$ independent vectors. In an $n$-dimensional space, that is a basis. A basis of eigenvectors is exactly what diagonalisation needs.\n",
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