605-Reply to a Discussion 3

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Eigenvalues

T10† Suppose that A is a square matrix. Prove that the constant term of the characteristic polynomial of A is equal to the determinant of A.

\[ A(x) = \sum_{0}^{n} a_{n}x^{n} \] For \(n=0, x = 0\) => \(A(0) = a_{0} = constant\)

Therefore \(\lambda = 0\),

=> \(A(0) = det(A - \lambda I_{n}) = det(A - 0 I_{n}) = det(A)\)