$A\succeq 0$, $A\in \mathbb{R}^{2\times 2}$. Are all expanded matrices produced by $A$ and $0$ positive semidefinite?

Question

I am not sure if this is a good title.

For any $A\succeq 0$, $A\in \mathbb{R}^{2\times 2}$. Consider the following expanded matrix:

$$M\in \mathbb{R}^{n\times n}$$

All entries of $M$ are zeros except for four entries:

$$M_{ii} = A(1,1),\quad M_{ij} = A(1,2), \quad M_{ji} = A(2,1),\quad M_{jj} = A(2,2)$$

where $1\le i<j\leq n$, for arbitrary $i$ and $j$.

Is $M$ still positive semidefiniteness?

Note: $A$ is symmetric. So by construction, $M$ is also symmetric.

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Since $A\succeq 0$, $A(1,1)\geq 0$, $A(2,2)\geq 0$ and $A(1,1)A(2,2)-A(1,2)^2\geq 0$. I think $M$ is PSD by Gershgorin's circle theorem.

Answer 1

\begin{align}x^TMx&=x^T\begin{bmatrix} 0 \\ M_{ii}x_i+M_{ij}x_j \\ 0 \\ M_{ji}x_i +M_{jj}x_j \\ 0\end{bmatrix} \\ &= x_iM_{ii}x_i+x_iM_{ij}x_j+x_jM_{ji}x_i+x_jM_{jj}x_j \\ &= \begin{bmatrix} x_i & x_j\end{bmatrix} A \begin{bmatrix} x_i \\x_j\end{bmatrix} \\ &\ge 0\end{align}