Can diagonally dominant matrix be written as the sum of diagonally dominant matrices?

Question

I am trying to work out an property of Eigen-decomposition.

Assume that

• $A$ is symmetric and all-positive, i.e. $A_{ij}=A_{ji}$ and $A_{ij}>0$.
• $A$ is diagonally dominant, i.e. $A_{ii}\geq \sum_{j\neq i}A_{ij}$ for all $i$.

Suppose the Eigen-decomposition of $A$ looks like $$A = U\Lambda U^T=\sum_i \lambda_iu_iu_i^T,$$ where $u_i$ is the $i$-th column of orthogonal $U$ and $\Lambda=\text{diag}(\lambda_1,...,\lambda_n)$.

Can we prove or disprove that (all or some) $u_iu_i^T$ are also diagonally dominant?

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What if we assume in addition that

• $A$ is normalized by its row/column sums (doubly stochastic matrix), i.e. $\sum_i A_{ij}=\sum_j A_{ij}=1$.

Answer 1

No, counterexample $$\begin{bmatrix}3&1\\1&2\end{bmatrix}$$ Which disprove the rule.