# Is it true that the determinant of symmetric positive definite matrix is the product of the eigenvalues?

I was working with a symmetric positive definite matrix when I encountered upon the following "identity"

Let A be symmetric pd

$\det(A)$

$= \det(Q\Lambda Q^{-1})$ (all symmetric matrices diag'able)

$= \det(Q)\det(\Lambda)\det(Q^{-1})$ (product of symmetric matrices)

$= \det(Q) \det(\Lambda) \dfrac{1}{\det{Q}}$ (determinant of inverse, property of orthogonal matrices)

$= \det(\Lambda)$ (determinant of diagonal matrix)

$= \prod\limits_{i=1}^n \lambda_i$

I looked for sometimes for a verification of this claim, but I did not come up with anything. Can anyone check that the claim is indeed correct or false?

• The claim is correct and $A$ don't have to be positive definite (Only need to be symmetric). – user99914 Sep 20 '15 at 13:25
• The determinant of any square matrix is the product of its eigenvalues. – Chris Godsil Sep 20 '15 at 13:25
• @ChrisGodsil Hi can you provide a reference, thanks – Carlos - the Mongoose - Danger Sep 20 '15 at 13:29
• @illegal immigrant: practically any linear algebra text - up to sign the constant term of the characteristic polynomial is the product of the eigenvalues. – Chris Godsil Sep 20 '15 at 13:33
• Just to be sure, by eigenvalue I need some $\lambda$ so that $Av = \lambda v$ for some nonzero $v$. If eigenvalues are defined as roots of the characteristic polynomials, then the statement is true for all $A$, as linked in the answer. – user99914 Sep 20 '15 at 14:06