From the description of positive semidefinite matrix, I found that determinant of positive semidefinite is multiplication of its eigenvalues. Can anyone show me how to derive this property?

  • $\begingroup$ Hint: Eigenvalue decomposition $\endgroup$ – user251257 Apr 4 '16 at 23:39

This holds for all matrices. Suppose that $\mathbf M \in \mathbb{F}^{n,n}$ is a matrix. Its eigenvalues $(\lambda_k)_{k=0}^{n-1}$ are roots of the characteristic polynomial $$ \det (\lambda \mathbf I - \mathbf M) = c \prod_{k=0}^{n-1} (\lambda_k - \lambda) $$ Expanding along the diagonal shows that the leading coefficient is $c = 1$. Therefore you can set $\lambda = 0$ to obtain the desired result.

  • $\begingroup$ Thank you for information. Can you explain me again the rationale behind setting c = 1 & λ=0? $\endgroup$ – pippp Apr 11 '16 at 11:15
  • $\begingroup$ @pippp Every polynomial can be factored, and the leading coefficient can vary. In this case the leading coefficient $c$ is $1$. We can set $\lambda = 0$ since $\lambda$ is an arbitrary variable. $\endgroup$ – Henricus V. Apr 11 '16 at 13:19

The determinant of any matrix $M$ is the constant term of the characteristic polynomial $\det(M - \lambda I)$, and thus the product of the roots of that polynomial, which are the eigenvalues (counted by algebraic multiplicity).

  • $\begingroup$ As long as the characteristic polynomial factorizes... $\endgroup$ – user251257 Apr 5 '16 at 10:05
  • $\begingroup$ Over a splitting field, it does. The OP mentioned "positive semidefinite", so in this case the field can be $\mathbb R$. $\endgroup$ – Robert Israel Apr 5 '16 at 15:18
  • $\begingroup$ yes. I think it is a fact worth to mention. $\endgroup$ – user251257 Apr 5 '16 at 16:39

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