# Positive semi-definite real matrix with unit diagonal

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 $$\det (M) = 0$$?

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

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

• "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? – Rodrigo de Azevedo Jul 23 '20 at 22:25
• I think you're right @RodrigodeAzevedo – TSF Aug 3 '20 at 14:58