1
$\begingroup$

Give an example of a $n\times n$ positive semi-definite real matrix $M\in \mathbb{R}^{n \times n}$, such that the following two conditions hold:

  1. the eigenvalues $\lambda_1, \dots, \lambda_n$ of $M$ are $\lambda_i \leq 1$ for all $i\in [n]$;

  2. the diagonal entries are $m_{i, i} = 1$, for all $i \in [n]$.

Is it possible to define any such matrix $M$ with the additional property that $\det (M) = 0$?

$\endgroup$
2
$\begingroup$

No.

Since we have a symmetric PSD matrix we have the following,

$$Tr(M) = \sum\limits_{i=1}^n \lambda_i$$

and

$$\det(M) = \prod\limits_{i=1}^n \lambda_i.$$

By assumption, $Tr(M) = \sum\limits_{i=1}^nm_{i,i}=\sum\limits_{i=1}^n 1= n$. Thus, $\sum\limits_{i=1}^n\lambda_i = Tr(M) = n$. Since, for each $i\in[n]$, $0\leq \lambda_i\leq 1$, we have that $\lambda_i=1$ for each $i\in[n]$. Then, the determinant is necessarily $1$ since

$$\det(M) = \prod\limits_{i=1}^n\lambda_i = \prod\limits_{i=1}^n 1 = 1.$$

$\endgroup$
2
  • $\begingroup$ "Since we have a symmetric PSD matrix..."? The trace and determinant do not follow from SPSD, but hold for every matrix. Or am I missing something? $\endgroup$ – Rodrigo de Azevedo Jul 23 '20 at 22:25
  • $\begingroup$ I think you're right @RodrigodeAzevedo $\endgroup$ – TSF Aug 3 '20 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.