I have the following matrix $A$:

  • symmetric
  • all positive and/or zero values
  • the main diagonal is all the same value, $x$.

To ensure that the matrix $A$, is positive semidefinite, must I only ensure that $x \geq 0$? It seems correct from my thinking, but wanted to make sure. Thanks.


2 Answers 2


It is not sufficient to have positive diagonal entries. To see this, consider the matrix $$ A=\pmatrix{1& 10\\10& 1}. $$ It has the negative eigenvalue $-9$ to the eigenvector $$ v=\pmatrix{1\\-1}, $$and is thus not positive semi-definite.


Note however that a diagonally dominant symmetric matrix is positive semi definite.

In your case it is sufficient to add the condition $x \ge \sum_{j\neq i} A_{ij}, i\neq j$ to ensure $A$ to be positive semi definite.

  • $\begingroup$ Why does it suffice for $x\ge A_{ij}$? I would appreciate a reference for this if you have one. $\endgroup$ Feb 28, 2018 at 1:25
  • $\begingroup$ @user3281410 The wikipedia article states this property for Hermitian diagonally dominant matrices, so it applies for real symmetric matrices as well : en.wikipedia.org/wiki/Diagonally_dominant_matrix $\endgroup$
    – Toool
    Mar 17, 2018 at 15:06
  • $\begingroup$ I believe you mean to say $x\ge \sum_{j\neq i}^{}A_{ij}$ for all $i$, not just $x\ge A_{ij}$ for all $i\neq j$. There are symmetric, matrices with nonnegative entries such that the entries on the diagonal are all the same (denoted $x$) and satisfy $x \ge A_{ij}$ but $A$ is not positive semidefinite. See the first answer here: math.stackexchange.com/questions/2669941/… $\endgroup$ Mar 20, 2018 at 22:56
  • $\begingroup$ @user3281410 Yep sorry I really meant $x \ge \sum_j A_{ij}, i\neq j$. I've updated the answer ! $\endgroup$
    – Toool
    Mar 21, 2018 at 14:36
  • $\begingroup$ @Tool is the converse statement true? If I have a positive semidefinite matrix, can I say that it must also be diagonally dominant with nonnegative real entries on the diagonal? $\endgroup$
    – Tyberius
    Mar 28, 2018 at 20:12

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