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

In the case that $A, B, C \in \mathbb{R}$, you can use the Sylvester's criterion for positive semi-definiteness. In short, this means any square sub-matrix along the diagonal must have a non-negative determinant. There are two cases we must consider:


*

*$A = 0$. Sylvester's Criterion tells us that $A, B, C \geq 0$ if the matrix $M$ is positive semi-definite (these are $1 \times 1$ matrices along the diagonal. We also know that $\det M \geq 0$ (this is a $3 \times 3$ matrix along the diagonal. Finally, looking at the upper left $2 \times 2$ matrix, we must have $BC - B^2 
\geq 0$. Noticing that $\det M = -C^2B \geq 0$, we conclude that if $A = 0$, then so must $B$ and $C$. 

*Let us assume that $A = 1$ (this is equivalent to all the cases where $A$ is nonzero), then looking at the bottom-right $2 \times 2$ matrix, we must have $B - 1 \geq 0$. Looking at the top left $2 \times 2$ matrix, we have $BC - B^2 \geq 0$. Since $B \geq 1$, this implies that $C \geq B$. Finally, we must have that the determinant of the entire matrix is non-negative: $$\det M=C(B - 1) - B(B - C) + C (B - BC) = -BC^2 + (3B - 1)C - B^2 \geq 0$$ This is a quadratic in $C$ with solution $C = \frac{1 - 3B \pm \sqrt{-4B^3 + 9B^2 - 6B + 1}}{-2B}$. We can notice that these solutions in $\mathbb{R}$ are necessarily negative when they exist with the exception of $B = 1$. However, we must have that $A, B, C \geq 0$, so we conclude that if $A = 1$, $B = 1$, which subsequently yields $C = 1$.
So, as @loup blanc observed, the only solutions where $M$ is positive definite are $A = B = C \geq 0$.
