Exercise 901Linear algebra › DeterminantsL1Added 3 Nov 2026

Determinant of a diagonalisable matrix

#diagonalisation#determinant#eigenvalues

Statement

Let $A$ be diagonalisable with eigenvalues $\lambda_1,\ldots,\lambda_n$ . Show that $\det A=\lambda_1\cdots\lambda_n$ .

Hint

Write $A=PDP^{-1}$ and use multiplicativity of the determinant.

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Solution

Diagonalisation. $A=PDP^{-1}$ with $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ .

Determinant.
$\det A = \det(PDP^{-1})=\det P\cdot\det D\cdot\det(P^{-1})=\det D=\prod_i\lambda_i.$

Conclusion. The determinant is the product of the eigenvalues (counted with multiplicity). This is also the constant term of the characteristic polynomial, up to sign.

Determinant of a diagonalisable matrix

Statement

Solution

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