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 positivedefinite? Notice that making $A$ positivesemidefinite 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 allone matrix.
1 Answer
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 $1$, $(n2)m$ entries equal to $0$), then we have $\langle Ax,x \rangle = 0$ and therefore $A$ cannot be positivedefinite.

$\begingroup$ 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

$\begingroup$ 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

$\begingroup$ Oh yes, tx, i read the sum of the diagonal of each block is one... $\endgroup$ May 29 at 13:28

$\begingroup$ Thanks a lot Guillaume! So this gives us that $0$ must be an eigenvalue with multiplicity at least $n1$. Is it possible that the multiplicity is exactly $n1$? $\endgroup$ May 29 at 13:55

$\begingroup$ Yes, I think that the matrix with diagonal blocks equal to $\frac{1}{m} I_m$ and offdiagonal blocks equal to $\frac{1}{m^2} J_m$ has this property. $\endgroup$ May 29 at 14:08