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  "body": "\n## Proof\n\nWe prove the formula for $P^{-1} A P$ first, and then both implications.\n\n1. Let $L \\colon F^n \\to F^n$ be the linear map $L(x) = A x$, and let $E = (e_1, \\dots, e_n)$ be the standard basis. Then $L(e_j) = A e_j$ is column $j$ of $A$, so the [[matrix of $L$|0YYMY95FWS]] in $E$ is $[L]_E = A$.\n2. Let $B' = (v_1, \\dots, v_n)$ be a [[basis|0VTQTPQGAV]] of $F^n$ with $A v_i = \\lambda_i v_i$ for each $i$. Let $P$ be the matrix with columns $v_1, \\dots, v_n$.\n3. Column $j$ of $P$ is $v_j = [v_j]_E$. So $P$ is the [[change of basis matrix|0KWXYZPJKN]] from $E$ to $B'$.\n4. Since $L(v_j) = \\lambda_j v_j$, column $j$ of $[L]_{B'}$ has $\\lambda_j$ in row $j$ and $0$ elsewhere. So $[L]_{B'} = D$, the [[diagonal matrix|0Q9PGXD9E4]] $\\operatorname{diag}(\\lambda_1, \\dots, \\lambda_n)$.\n5. By the [[change of basis formula|0MQXVZEBMT]], $P$ is invertible and $[L]_{B'} = P^{-1} [L]_E P$. By steps 1 and 4, $P^{-1} A P = D$.\n6. This proves the formula of the statement. It also shows that if $F^n$ has a basis of eigenvectors of $A$, then $A$ is [[diagonalisable|0Y17PEKF24]].\n7. Conversely, let $A$ be diagonalisable. So there is an [[invertible|0TWTS88TWN]] matrix $P$ with $P^{-1} A P = D = \\operatorname{diag}(d_1, \\dots, d_n)$.\n8. Let $v_1, \\dots, v_n$ be the columns of $P$. Multiplying by $P$ on the left gives $A P = P D$.\n9. Column $j$ of $A P$ is $A v_j$. Column $j$ of $P D$ is $P (d_j e_j) = d_j v_j$. So $A v_j = d_j v_j$ for each $j$.\n10. Let $c_1, \\dots, c_n \\in F$ with $\\sum_j c_j v_j = 0$, and let $c$ be the column with entries $c_j$. Then $P c = 0$, so $c = P^{-1} P c = 0$.\n11. By step 10, $v_1, \\dots, v_n$ are [[linearly independent|0EVHJFTG9Z]]. The space $F^n$ has [[dimension $n$|0QK7FPE6VE]], so they form a basis [[by counting|096V7H13H9]].\n12. Each $v_j$ is nonzero, since it belongs to a basis. By step 9, $v_j$ is an [[eigenvector|0B5H4208Z9]] of $A$ for $d_j$.\n13. So $F^n$ has a basis of eigenvectors of $A$. Together with step 6, this proves the equivalence.\n\n## Idea\n\nThe equation $A P = P D$ read column by column says exactly that each column of $P$ is an eigenvector, with eigenvalue the matching diagonal entry of $D$. The matrix $P$ is invertible exactly when its columns form a basis. So diagonalising $A$ and finding a basis of eigenvectors are the same task.\n",
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