How to get roots of determinant (characteristic) equation?

Hello all, I am solving an eigenvalue problem and giving symbolic matrix as input. I want to find roots of characteristic equation, I mean, roots of determinant of matrix equated to zero. Here I give script:

clear all;
close all;
clc;

syms w
A=[-2000*w^2+280*1e3,-280*1e3;280*1e3,-2000*w^2+280*1e3];

fun = matlabFunction(det(A))

I want to find roots of fun(). This is a polynomial equation of 4th order, so I should have 4 roots. If I use fzero, it just gives a local solution to problem, but I want to have all roots. Can you suggest something? Ofcourse, I can write coefficients of det(A) manually and pass it to roots([...]). But I don't want to write manually. I am even trying to bypass symbolics, as for large matrix, symbolic variables are computationally very expensive. Any comments? Thanks in advance!

Answer 1

[Star Strider]

Use the Symbolic Math Toolbox solve function:

DA = det(A)
W = solve(DA,w)

produces:

W =

    (140 + 140*i)^(1/2)
    (140 - 140*i)^(1/2)
   -(140 + 140*i)^(1/2)
   -(140 - 140*i)^(1/2)

[Amit Kumar]

Thanks a lot!! This solve() I missed. Anyway, is there any way to avoid symbolics, by creating say a function of w? Symbolics are computationally slow, especially for large matrices.

[Star Strider]

Found it!

p = sym2poly(DA)

r = roots(p)

produces:

p =

       4.0000e+006     0.0000e+000    -1.1200e+009     0.0000e+000   156.8000e+009

r =

     -12.9998e+000 +  5.3847e+000i
     -12.9998e+000 -  5.3847e+000i
      12.9998e+000 +  5.3847e+000i
      12.9998e+000 -  5.3847e+000i

[Amit Kumar]

Thanks a lot!!!

[Star Strider]

My pleasure!

Answer 2

[Pratik Baraiya]

clear all;
close all;
clc;
syms w
A=[-2000*w^2+280*1e3,-280*1e3;280*1e3,-2000*w^2+280*1e3];
fun = matlabFunction(det(A))

fun = function_handle with value:
    @(w)w.^2.*-1.12e+9+w.^4.*4.0e+6+1.568e+11

[Walter Roberson]

What is your recommendation to proceed from fun to find the roots of fun ?