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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?

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Here's a MathJax tutorial :) – Shaun Jul 6 '14 at 20:04
Welcome to MSE! You should try to share with us your thoughts, what you can assume (theorems, lemmata, etc, you know), to make it more likely to get good help. – gnometorule Jul 6 '14 at 20:13

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.

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Very nice derivation. This also shows why you cannot maintain positive definiteness. – Jonas Dahlbæk Jul 6 '14 at 20:20
thank you. that's very helpful – Gal Jul 6 '14 at 22:34
@Gal: If you are satisfied with the answer, you can accept it by clicking the check mark on the left. – Rahul Jul 7 '14 at 17:29

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.

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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 $$\langle z,Bz\rangle=\left\langle\left( \begin{array}{cc} x \\ y \\ \end{array} \right),\left( \begin{array}{cc} Ax+Ay \\ Ax+Ay \\ \end{array} \right)\right\rangle=xAx+xAy+yAx+yAy=(x+y)A(x+y)\geq0$$

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thank you. that's very helpful. That's similar to the first answer – Gal Jul 6 '14 at 22:35
Did take me some time to write it down, so I didn't see that there was some similiar post. Rahuls answer is a bit more general – Michael Jul 7 '14 at 14:57

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.

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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.

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Positive semi-definite means $\langle Ax,x\rangle\geq 0$ for all $x$. – Jonas Dahlbæk Jul 6 '14 at 19:59

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