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  "body": "\n## Statement\n\nLet $K$ be a commutative ring, let $A$ be an $n \\times n$ matrix over $K$, and let $P$ be an invertible $n \\times n$ matrix over $K$.\nThen the [[characteristic polynomials|0VSSPAFM2C]] of $A$ and $P^{-1} A P$ are equal:\n$$p_{P^{-1} A P}(t) = p_A(t).$$\nSo [[similar matrices|0E89HQATM8]] have the same characteristic polynomial.\n\n## Remarks\n\nThe proof is $\\det(t I - P^{-1} A P) = \\det(P^{-1} (t I - A) P) = \\det(t I - A)$.\nHence the characteristic polynomial of a linear map does not depend on the basis.\nThe converse fails: $I_2$ and $\\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 1 \\end{pmatrix}$ both have $p(t) = (t - 1)^2$, but they are not similar.\n\n## Explanations\n\n### L1 human\n\nA change of basis does not change the characteristic polynomial.\nSo the eigenvalues, the trace and the determinant, which we can read from it, are the same in every basis.\nThey are properties of the linear map, not of one matrix.\n",
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