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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?

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  • $\begingroup$ The claim is correct and $A$ don't have to be positive definite (Only need to be symmetric). $\endgroup$ – user99914 Sep 20 '15 at 13:25
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    $\begingroup$ The determinant of any square matrix is the product of its eigenvalues. $\endgroup$ – Chris Godsil Sep 20 '15 at 13:25
  • $\begingroup$ @ChrisGodsil Hi can you provide a reference, thanks $\endgroup$ – Carlos - the Mongoose - Danger Sep 20 '15 at 13:29
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    $\begingroup$ @illegal immigrant: practically any linear algebra text - up to sign the constant term of the characteristic polynomial is the product of the eigenvalues. $\endgroup$ – Chris Godsil Sep 20 '15 at 13:33
  • $\begingroup$ 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. $\endgroup$ – user99914 Sep 20 '15 at 14:06
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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

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    $\begingroup$ Wow so it is for every matrix. I don't know why I am shocked because I probably derived this a thousand times before but just seeing it used in a completely different context (convex optimization) has led me to believe it is some work of ingenuity of the optimization field not linear algebra $\endgroup$ – Carlos - the Mongoose - Danger Sep 20 '15 at 14:04

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