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

Question

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?

Answer 1

Yes the claim is indeed true. The determinant of a matrix is equal to the product of its eigenvalues. Check this link determinant is equal to the product