0

I need to get a Diagonalized of a matrix. My given matrix is 4*4 strictly Dominant Diagonal matrix. something like this: \begin{array}{l}4&2&1&0\\1&5&2&1\\2&0&6&3\\1&1&0&3\end{array} it seems the result of Diagonalized of matrix ( if it is possible) would be in form of A= S * D * S_inverse. So now I really need to have the D matrix as all entries are integers not fractional. is there any way or any condition that A as strictly Dominant Diagonal matrix can have D as diagonal matrix with all integers? I used this website for matrix calculation, it is hard to just try infinite numbers to get finally your desired output.
calculator website

EDITED strictly Dominant Diagonal matrix : a matrix which items on diagonal is strictly greater than absolute sum of all items in that row, means a_ii > Sum ( | a_ij | ) for all j, i != j as you see in my example 4 > 2 +1 + 0 and 5 > 1 + 2 + 1 and so on

Azzurro94
  • 101
  • What is "strictly dominant diagonal"? Is it the same as here? – Dietrich Burde Aug 06 '21 at 18:23
  • @DietrichBurde yes, please see my edited also – Azzurro94 Aug 06 '21 at 18:50
  • If a matrix is similar to a diagonal matrix $D,$ the diagonal entries of $D$ are precisely the eigenvalues. For your matrix, those are irrational, not integers – Will Jagy Aug 06 '21 at 19:02
  • I know my matrix is just an example. I need to find a A matrix which D would be integers as well. any way I found one by lots of trial but I found there are a theorem. – Azzurro94 Aug 06 '21 at 20:27
  • 1
    If you are looking for examples, surely you should do it backwards. Pick a $D$, $S$ and just compute $A$. Here is an example with all integers: $A = \begin{bmatrix}6 & 2 & -2\ 1 & 7 & -1\ -1 & 1 & 5\end{bmatrix}$ which has eigenvalues 4, 6, 8. – Mikael Öhman Aug 06 '21 at 21:08

0 Answers0