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
 A: If the blocks have dimension $1$, then, the sole solutions are $A=B=C\geq 0$.
EDIT. That follows is a solution that uses , like that of @user1551, the above remark.
We consider the matrix $M=U\otimes C+V\otimes B+W\otimes A$, where $U=\begin{pmatrix} 1&0&1\\ 0&0&0\\ 1&0&0\end{pmatrix},\cdots$. Then $(x\otimes y)^TM(x\otimes y)\geq 0$ for every vectors $x\in\mathbb{C}^3,y\in\mathbb{C}^k$, that is, 
$x^TUx(y^TCy)+x^TVx(y^TBy)+x^TWx(y^TAy)\geq 0$; thus, for every $x$ and for a fixed $y$,
$x^T(uU+vV+wW)x\geq 0$, where $u=y^TCy,\cdots$. The associated matrix 
$\begin{pmatrix}u&v&u\\v&v&w\\u&w&w\end{pmatrix}$ must be $\geq 0$, that happens only if $u=v=w$.
Finally, for every $y$, $y^TCy=y^TBy=y^TAy$, that implies that $A=B=C$.
A: 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 $A = B = C \geq 0$.
A: 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 $A=B=C$.
