How many positive elements in a diagonal matrix $D$ do we need to make $A+D$ positive definite for a real symmetric $A$ with $m$ negative eigenvalues

Question

Consider a $n$-dimensional real symmetric matrix $A\in\mathbb{R}^{n\times n}$ having exactly $m$ negative eigenvalues and $n-m$ positive eigenvalues ($1\leq m\leq n$). Let $D\in\mathbb{R}^{n\times n}$ be a diagonal matrix with non-negative elements. My question is: To make $A+D\succ0$, i.e. positive definite, at least how many positive elements of $D$ do we need? For $n=1, 2$, it is easy to show that the answer is exactly $m$. Does this conclusion still hold for $n\geq3$ ?

Any ideas and suggestions are welcomed. Thanks in advance!

Answer 1

No, it's not necessarily $m$. Consider the matrix

$$ A = \pmatrix{1 & -1 & -1\cr -1 & 1 & -1\cr -1 & -1 & 1\cr} $$ This has $m=1$ negative eigenvalue (namely $-1$), but if $D$ is a diagonal matrix with one nonzero element, $\det(A+D) = \det(A) = -4$ because the cofactor is $0$.

Answer 2

For $n=1, 2$, it is easy to show that the answer is exactly $m$.

This is not true. When $m>0$, the answer is $n$, because there exists some $A$ with a negative diagonal.

For instance, pick any entrywise positive vector $u_1$ and complete it to an orthonormal basis $\{u_1,u_2,\ldots,u_n\}$ of $\mathbb R^n$. For any $c>0$, the matrix $A=-cu_1u_1^T-\sum_{1<j\le m}^nu_ju_j^T+\sum_{j>m}u_ju_j^T$ has $m$ negative eigenvalues and $n-m$ positive eigenvalues, but it is entrywise negative when $c$ is sufficiently large.

In particular, such an $A$ has a negative diagonal. If the diagonal of $D$ is not entirely positive, $A+D$ will have at least one negative diagonal element and hence it cannot be positive definite.

Edit. To construct a nice numerical example, use matrix congruence rather than orthogonal similarity. E.g. let $c>1$. The matrix $$ A=-cee^T-\sum_{1<j\le m}^ne_je_j^T+\sum_{j>m}e_je_j^T $$ with all off-diagonal entries equal to $-c$ and with a negative diagonal consisting of one instance of $-c$, $m-1$ copies of $-c-1$ and $n-m$ copies of $-c+1$, has exactly $m$ negative eigenvalues and $n-m$ positive eigenvalues because it is congruent to $(-c)\oplus(-I_{m-1})\oplus I_{n-m}$.