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  "body": "\n## Statement\n\nLet $F$ be a field and let $A$ be an $n \\times n$ [[square matrix|0JXZKX526R]] over $F$.\nSuppose that $A$ has $n$ [[linearly independent|0CRMGPAH83]] [[eigenvectors|0N6960W0BV]] $q_1, \\dots, q_n$ in $F^n$, with $A q_i = \\lambda_i q_i$.\nLet $Q$ be the matrix with columns $q_1, \\dots, q_n$, and let $\\Lambda$ be the [[diagonal matrix|044JNC7ZAA]] with diagonal entries $\\lambda_1, \\dots, \\lambda_n$.\nThen $Q$ is [[invertible|0EB9N0S33C]] and\n$$A = Q \\Lambda Q^{-1}.$$\nConversely, let $A = Q \\Lambda Q^{-1}$ with $Q$ invertible and $\\Lambda$ diagonal.\nThen the columns of $Q$ are linearly independent eigenvectors of $A$.\nThe diagonal entries of $\\Lambda$ are the corresponding [[eigenvalues|0BQAA2BK3Q]].\n\n## Remarks\n\nThe columns of $AQ$ are the vectors $A q_i = \\lambda_i q_i$, and these are also the columns of $Q \\Lambda$.\nThus $AQ = Q\\Lambda$, and $Q$ is invertible because its columns are independent.\nFor example, $A = \\begin{pmatrix} 2 \u0026 1 \\\\ 0 \u0026 3 \\end{pmatrix}$ has the eigenvector $(1, 0)^T$ for $2$ and $(1, 1)^T$ for $3$.\nThus $A = Q \\Lambda Q^{-1}$ with $Q = \\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 1 \\end{pmatrix}$ and $\\Lambda = \\operatorname{diag}(2, 3)$.\nThe hypothesis is necessary: $\\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$ has only one independent eigenvector, and it has no eigendecomposition.\nWhen $A$ is real symmetric or complex [[normal|0MR0BYF0K2]], $Q$ can be chosen orthogonal or unitary, by the [[spectral theorem|06J4PXRG22]].\n\n## Explanations\n\n### L1 human\n\nA matrix with enough eigenvectors acts, in the basis of those eigenvectors, by stretching each axis by its eigenvalue.\nThe eigendecomposition writes this as: change of basis, stretch, change back.\nNot every matrix has one; exactly the [[diagonalisable|07SQCNV233]] matrices do.\n\n### L2 human\n\nThe decomposition makes powers and functions of $A$ easy: $A^k = Q \\Lambda^k Q^{-1}$, and the same holds for the exponential.\nIt is the matrix form of the criterion that [[a matrix is diagonalisable iff it has a basis of eigenvectors|056SW1E0AC]].\nDistinct eigenvalues give enough eigenvectors, by [[this statement|0HNMFG8F8K]].\nWhen eigenvectors are missing, the [[Jordan normal form|085HQ5G5TE]] or the [[Schur decomposition|V1JG4C7R7B]] takes its place.\n",
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