# Positive-definite block matrix with constant block sums

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.

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.
• This not a general rule say $A=1/3\begin{pmatrix}1&1&1.5&0\\1&2&0&1.5\\1.5&0&1&0\\0&1.5&0&2\end{pmatrix} (n=m=2)$? May 29 at 12:58
• This $A$ does not satisfy the hypothesis since the sum of the entries in the northwest block is $2$ and not $1$. May 29 at 13:21
• 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 $n-1$? May 29 at 13:55
• 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. May 29 at 14:08