Given two natural numbers $n$ and $m$, suppose that $A$ is an $nm \times nm$ real nonnegative matrix. Seeing $A$ as a block matrix where each block has size $m\times m$, suppose that the sum of the entries in each block is $1$. Can $A$ be positive-definite? Notice that making $A$ positive-semidefinite is easy, for example we can let $A=\frac{1}{m}J_n\otimes I_m$. Also, notice that for $m=1$ the answer is no, as the only matrix satisfying the constraints is the all-one matrix.
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This is not possible unless $n=1$. If $x$ is the vector $$ (1,\dots,1,-1,\dots,-1,0,\dots,0) $$ ($m$ entries equal to $1$, $m$ entries equal to $-1$, $(n-2)m$ entries equal to $0$), then we have $\langle Ax,x \rangle = 0$ and therefore $A$ cannot be positive-definite.
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$\begingroup$ This not a general rule say $A=1/3\begin{pmatrix}1&1&1.5&0\\1&2&0&1.5\\1.5&0&1&0\\0&1.5&0&2\end{pmatrix} (n=m=2)$? $\endgroup$ May 29 at 12:58
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$\begingroup$ This $A$ does not satisfy the hypothesis since the sum of the entries in the northwest block is $2$ and not $1$. $\endgroup$ May 29 at 13:21
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$\begingroup$ Oh yes, tx, i read the sum of the diagonal of each block is one... $\endgroup$ May 29 at 13:28
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$\begingroup$ Thanks a lot Guillaume! So this gives us that $0$ must be an eigenvalue with multiplicity at least $n-1$. Is it possible that the multiplicity is exactly $n-1$? $\endgroup$ May 29 at 13:55
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$\begingroup$ Yes, I think that the matrix with diagonal blocks equal to $\frac{1}{m} I_m$ and off-diagonal blocks equal to $\frac{1}{m^2} J_m$ has this property. $\endgroup$ May 29 at 14:08