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Every square matrix satisfies its own characteristic equation.
If $$\det(A - \lambda I) = (-1)^n \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_0 = 0$$, then $$(-1)^n A^n + c_{n-1}A^{n-1} + \cdots + c_0 I = 0$$.
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Every square matrix satisfies its own characteristic equation.
If $$\det(A - \lambda I) = (-1)^n \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_0 = 0$$, then $$(-1)^n A^n + c_{n-1}A^{n-1} + \cdots + c_0 I = 0$$.
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