Positive semi-definite real matrix with unit diagonal

Question

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

the eigenvalues $lambda_1, dots, lambda_n$ of $M$ are $lambda_i leq 1$ for all $iin [n]$;

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 $det (M) = 0$?

TSF · Answer

No.
Since we have a symmetric PSD matrix we have the following,
$$Tr(M) = sumlimits_{i=1}^n lambda_i$$
and
$$det(M) = prodlimits_{i=1}^n lambda_i.$$
By assumption, $Tr(M) = sumlimits_{i=1}^nm_{i,i}=sumlimits_{i=1}^n 1= n$. Thus, $sumlimits_{i=1}^nlambda_i = Tr(M) = n$. Since, for each $iin[n]$, $0leq lambda_ileq 1$, we have that $lambda_i=1$ for each $iin[n]$. Then, the determinant is necessarily $1$ since
$$det(M) = prodlimits_{i=1}^nlambda_i = prodlimits_{i=1}^n 1 = 1.$$

Positive semi-definite real matrix with unit diagonal

One Answer

Add your own answers!

Ask a Question