3

If I have a $n$-dim matrix $A=\{a_{ij}\}$, and I multiply each elements by a factor $w_{ij}$ in $[0,1]$, and get a new matrix $A_w=\{a_{ij}w_{ij}\}$. Do I have $$||A||\ge \lVert A_w\rVert$$ where the norm is the operator norm, i.e. the largest eigenvalue of $A$?

If not, what if we add some conditions, say,

  • $A$ is elementwise positive
  • $A$ is PSD
  • $A$ is symmetric
  • $w_{ij}$ is 1 on major diagonals, and decreases to 0 gradually to the northeast, and southwest corner?

Thanks!

Davide Giraudo
  • 181,608
breezeintopl
  • 1,477

1 Answers1

3

In general, no. Consider $A=1\oplus\pmatrix{1&1\\ 1&-1}\oplus1$ and $W=1\oplus\pmatrix{1&0\\ 1&1}\oplus1$. We have $\|A\|_2=\sqrt{2}<\frac{1+\sqrt{5}}2=\|A_w\|_2$.

As for your proposed conditions, here is a partial answer:

  • When $A$ is entrywise nonnegative, $A^TA\ge A_w^TA_w\ge0$ entrywise. Hence the same order is preserved for their induced 1-norms and in turn $\|A\|_2^2=\rho(A^TA)\ge \rho(A_w^TA_w)=\|A_w\|_2^2$, by Gelfand's formula. (In general, the spectral radius of a positive matrix increases with its entries' sizes.)
  • (Edit.) The condition that $A$ is positive semidefinite does not work. You may see a nice counterexample here.
  • The mere condition that $A$ is symmetric is not enough. See the above counterexample.
  • The mere condition that the entries of $W$ decreases from 1 on the diagonal to 0 on the boundary is not effective. Again, see the above counterexample.
Community
  • 1
user1551
  • 149,263