0
$\begingroup$

Let $\lambda$ and $\Lambda$ be two fixed positive numbers and let $A$ be a symmetric real matrix with $\lambda |x|^2 \leq (Ax,x) \leq \Lambda |x|^2$.

Let $B$ be a matrix derived from $A$ as follows: $B$ retains the diagonal entries of $A$ and one off diagonal entry. An example would be $B$ having the diagonal and one above and below the diagonal entries from $A$ and the rest to be $0$.

Another example would be $B$ having diagonal and two above and below the diagonal entries from $A$. $$ \left(\begin{array}{ccccc}a_{11} & 0 & 0 & \ldots & 0\\ 0 & a_{22} & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & a_{nn} \end{array} \right), \left(\begin{array}{ccccc}a_{11} & a_{12} & 0 & \ldots & 0\\ a_{21} & a_{22} & a_{23} & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & a_{nn} \end{array} \right), \left(\begin{array}{ccccc}a_{11} & 0 & a_{13} & \ldots & 0\\ 0 & a_{22} & 0 & \ldots & 0 \\ a_{31} & 0 & a_{33} & \ldots & 0\\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & a_{nn} \end{array} \right) $$

A bunch of such matrices can be constructed and are clearly symmetric. Is it also true that these matrices satisfy bounds of the form $\lambda |x|^2 \leq (Ax,x) \leq \Lambda |x|^2$ with the same $\lambda$ and $\Lambda$? Clearly the diagonal matrix does, but what about the others?

$\endgroup$
2
  • 1
    $\begingroup$ To be clear, the details here matter a lot. Iosif Pinelis's answer shows that the answer is "no" if you can retain 2 or more non-zero off-diagonal entries above the diagonal (like in your second example?). But if you can only retain 1 non-zero off-diagonal entry (as specified in your text?) then you just have a bunch of $2 \times 2$ and $1 \times 1$ blocks on the diagonal, so the answer is "yes" (by pinching/interlacing inequalities or something like that). $\endgroup$ Commented May 2 at 18:17
  • $\begingroup$ I was specifically looking for matrices that retain both the bounds and symmetry. Are there other possibilities besides the diagonal matrix that can preserve both? $\endgroup$
    – Adi
    Commented May 3 at 18:17

1 Answer 1

1
$\begingroup$

A counterexample: $$A=\left( \begin{array}{ccc} 4 & 4 & -4 \\ 4 & 6 & -5 \\ -4 & -5 & 6 \\ \end{array} \right),\quad B=\left( \begin{array}{ccc} 4 & 4 & 0 \\ 4 & 6 & -5 \\ 0 & -5 & 6 \\ \end{array} \right).$$ Then $A$ is positive definite but $B$ is not.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .