Sort of positive matrix Hy i want to find a block  real or complex matrix of the form $\begin{pmatrix}C&B&C\\B&B&A\\C&A&A\end{pmatrix} $ such that it will be positive semi-definite but not such $A=B=C.$ All blocks are of same size and hermitian.
I couldn't find such a matrix in dimension three so that's why i am asking
Thanks

Call your matrix $M$. If $A,B,C$ are scalars, Sylvester's criterion shows that $M$ is PSD if and only if $A=B=C\ge0$. Now, suppose $A,B,C$ are $n\times n$ Hermitian matrices for some $n\ge2$ and $M$ is PSD. Then for each $i\in\{1,2,\ldots,n\}$, the principal submatrix $\pmatrix{c_{ii}&b_{ii}&c_{ii}\\ b_{ii}&b_{ii}&a_{ii}\\ c_{ii}&a_{ii}&a_{ii}}$ is also PSD. Therefore, using the result for the scalar case, we obtain $a_{ii}=b_{ii}=c_{ii}=$ some $d_i\in\mathbb R$.
Similarly, if we pick any $i,j\in\{1,2,\ldots,n\}$ with $i\ne j$, the principal submatrix
$$
\pmatrix{
m_{ii}&m_{ij}&m_{i,n+j}\\
m_{ji}&m_{jj}&m_{j,n+j}\\
m_{n+j,i}&m_{n+j,j}&m_{n+j,n+j}}
=\pmatrix{
c_{ii}&c_{ij}&b_{ij}\\
c_{ji}&c_{jj}&b_{jj}\\
b_{ji}&b_{jj}&b_{jj}}
=\pmatrix{
d_i&c_{ij}&b_{ij}\\
c_{ji}&d_j&d_j\\
b_{ji}&d_j&d_j}
$$
is PSD. However, as the last expression above is congruent to
$$
X=\pmatrix{1\\ &1&0\\ &-1&1}
\pmatrix{
d_i&c_{ij}&b_{ij}\\
c_{ji}&d_j&d_j\\
b_{ji}&d_j&d_j}
\pmatrix{1\\ &1&-1\\ &0&1}
=\pmatrix{
d_i&c_{ij}&b_{ij}-c_{ij}\\
c_{ji}&d_j&0\\
b_{ji}-c_{ji}&0&0},
$$
by Sylvester's criterion we must have $\det\pmatrix{x_{11}&x_{13}\\ x_{31}&x_{33}}=-|b_{ij}-c_{ij}|^2\ge0$, i.e. $b_{ij}=c_{ij}$.
Finally, as $M$ also contains the principal submatrix
$$
Y=\pmatrix{
m_{ii}&m_{i,n+i}&m_{i,2n+j}\\
m_{n+i,i}&m_{n+i,n+i}&m_{n+i,2n+j}\\
m_{2n+j,i}&m_{2n+j,n+i}&m_{2n+j,2n+j}}
=\pmatrix{
c_{ii}&b_{ii}&c_{ij}\\
b_{ii}&b_{ii}&a_{ij}\\
c_{ji}&a_{ji}&a_{jj}}
=\pmatrix{
d_i&d_i&c_{ij}\\
d_i&d_i&a_{ij}\\
c_{ji}&a_{ji}&d_j},
$$
a similar argument to the above shows that we also have $a_{ij}=c_{ij}$. Thus $A=B=C$.
