# Symmetric block matrices with zero trace

In my research in graph theory i am getting symmetric matrices with trace zero of this kind

$$\begin{bmatrix} 0 & 1 & 2 & 3 & 3 \\ 1 & 0 & 1 & 2 & 2 \\2 & 1 & 0 & 1 & 1 \\ 3 & 2 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 0 \end{bmatrix}$$ In general $$D_{M} (G) = \begin{bmatrix} 0_{k\times k} & 1_{k\times k_{1}} & 2_{k\times k_{2}} & . . . & d_{k\times k_{d}}\\ 1_{k_{1}\times k} & 0_{k_{1}\times k_{1}} & 1_{k_{1}\times k_{2}} & . . . & (d-1)_{k_{1}\times k_{d}} \\ 2_{k_{2}\times k} & 1_{k_{2}\times k_{1}} & 0_{k_{2}\times k_{2}} & . . . & (d-2)_{k_{2}\times k_{d}}\\ . & . & . & . . . & .\\ . & . & . & . . . & .\\ . & . & . & . . . & .\\ d_{k_{d}\times k} & (d-1)_{k_{d}\times k_{1}} & (d-2)_{k_{d}\times k_{2}} & . . . & 0_{k_{d}\times k_{d}} \end{bmatrix}$$ Here the entries are all non-negative integers and $i_{m\times n}$ denotes a block whose entries are all i's 1. I find the matrices to be singular, can that be proved generally? 2. Also the numerical value of the largest eigenvalue increases with the increase in the value of even one entry, how to prove this?

For your first question, what you describe does not always hold. The easiest counterexample is $\det\pmatrix{0&1\\ 1&0}\ne0$. However, if $k>1$ or some $k_i>1$, the matrix is obviously singular because it has two identical rows.

As for your second question, if the matrix is zero, the assertion is obvious; if $d\ge1$, the assertion is still true because your matrix is entrywise nonnegative and irreducible (the matrix is primitive when $d\ge2$ and its entries are flipping between $0$ and $1$ when $d=1$). See this thread for detailed proofs.

• user 1551 the given matrix can never be primitive because the diagonal elements are always zero.pl clarify\\
– Anu
Commented May 26, 2016 at 14:34
• user 1551 ,the given matrix can never be primitive because the diagonal elements are always zero.pl clarify \\ also can we compare the energy (the sum of the abs values of the eigenvalues) of two such matrices if one matrix has greater value of $d$ than the other
– Anu
Commented May 26, 2016 at 14:42
• @Anu An entrywise nonnegative matrix $A$ is called primitive if its powers are eventually entrywise positive, i.e. if $A^k$ is entrywise positive for some $k>0$. Your matrix is primitive because its square is entrywise positive. I'm not sure if one can cook up a proof using the "energy" (technically, the trace norm), but I would expect such a proof to be indirect, as the the trace norm in general is not a monotonic function of the spectral radius. Commented May 26, 2016 at 15:13
• 1551 do you mean by $A^k$ is entrywise positive to mean nonnegative because any power of $A$ will have diagonal entries zero Can you pl clarify?
– Anu
Commented May 27, 2016 at 12:06
• @Anu "... because any power of $A$ will have diagonal entries zero..." Huh? What are you talking about? Could you calculate the square of the first matrix in your question? Commented May 28, 2016 at 1:04