Positive semi-definite of a matrix composed of semi-definite blocks Say a matrix A is positive semi-definite. Let B be a square matrix composed of replicas of A as sub-blocks, s.t. $$B=\begin{pmatrix} A & A \\ A & A \\ \end{pmatrix},$$ or $$\begin{pmatrix} A & A & A \\  A & A & A \\ A & A & A \\ \end{pmatrix},$$ etc. Would $B$ be semi-definite as well?
 A: For any vector $x$, divide it into appropriately-sized subvectors $x_1,\ldots,x_n$ so that
$$\begin{align}
x^TBx &= \begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}^T\begin{bmatrix}A&\cdots& A\\\vdots&\ddots&\vdots\\A&\cdots&A\end{bmatrix}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} \\
&= x_1^TAx_1 + \cdots + x_1^TAx_n \\
&\phantom{=}+ \cdots \\
&\phantom{=}+ x_n^TAx_1 + \cdots + x_n^TAx_n \\
&= (x_1+\cdots+x_n)^TA(x_1+\cdots+x_n)
\end{align}$$
which is nonnegative because $A$ is positive semidefinite.
A: 
Let $X$ and $Y$ be symmetric positive semidefinite matrices. Then $X\otimes Y$, where $\otimes$ denotes the Kronecker product, is symmetric positive semidefinite as well.

Let
$$
X=E, \quad Y=A,
$$
where $E$ is a square matrix of ones (for the matrices in question, $E$ is $2\times 2$ or $3\times 3$). Note that $E=ee^T$ with $e=[1,1,\ldots,1]^T$, so $E$ is obviously symmetric positive semidefinite. Now use the fact above.
A: No, consider the 2-by-2 matrix with 1 in all entries.
Edit: I may have been too quick here, that matrix so defined actually is positive semi-definite, but it is not positive definite.
Edit2: I think you are correct: Let $A\geq 0$ and let $M$ be the '2-by-2' block matrix with $A$ in all 4 blocks. Then
$$
\langle M(x,y),(x,y)\rangle = \langle Ax,x\rangle +\langle Ay,y\rangle +\langle Ax,y\rangle+\langle Ay,x\rangle=\langle Ax,x\rangle +\langle Ay,y\rangle +2Re \langle Ax,y\rangle.
$$
Now, since $A\geq 0$, we can employ the Cauchy-Schwarz inequality to obtain
$$
|Re \langle Ax,y\rangle|\leq |\langle Ax,y\rangle|\leq \sqrt{\langle Ax,x\rangle\langle Ay,y\rangle}\leq \frac{1}{2}(\langle Ax,x\rangle + \langle Ay,y\rangle),
$$
which implies that $M\geq 0$. I believe this will go through to higher orders as well.
A: Let $$B=\left(
\begin{array}{cc}
 A & A \\
 A & A \\
\end{array}
\right)$$
Decomposit an Vector $z\in\mathbb{R}^{2n}$ to
$$z=\left(
\begin{array}{cc}
 x  \\
 y  \\
\end{array}\right)\quad x,y\in \mathbb{R}^n$$
then
$$Bz=\left(
\begin{array}{cc}
 A & A \\
 A & A \\
\end{array}
\right)
\left(
\begin{array}{cc}
 x  \\
 y  \\
\end{array}
\right)=\left(
\begin{array}{cc}
 Ax+Ay  \\
 Ax+Ay  \\
\end{array}
\right)
$$
So
$$\langle z,Bz\rangle=\left\langle\left(
\begin{array}{cc}
 x  \\
 y  \\
\end{array}
\right),\left(
\begin{array}{cc}
 Ax+Ay  \\
 Ax+Ay  \\
\end{array}
\right)\right\rangle=xAx+xAy+yAx+yAy=(x+y)A(x+y)\geq0$$
A: Yes in special cases because the determinant of a block matrix is the product of the
 determinants of the blocks, if the blocks are placed in the the main diagonal. This 
 is not true for general block matrices. Something like 
[A 0 0]
[0 B 0]
[0 0 C]
is allowed.
