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 $\det (M) = 0$?



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

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


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

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

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

  • $\begingroup$ "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

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