Eigenvalue doesn't zero the characteristic polynomial

I created a symmetric 21 x 21 matrix A with condition number 7.5044. I wrote the following code:

    syms x;
    AS = sym(A);
    polyA = charpoly(AS,x);
    solve(polyA);

and This is the result

          93.988842796147505104718718233034
    63.318989852350135397493503048259
    38.392863684000190231816927188875
    57.023973670819117411135094232301
    20.964197497113106525263127789941
    52.100859003607712279511955297751
    14.457776205660758291480361189262
    72.543368263184623095490492872815
   -19.500133944896490949432003697463
    65.733078713316290701540791543043
    71.723898977718034735792925663278
    57.262150118578505531059649955477
    27.620774843148443240592651344762
    49.356126883789838887152595264487
    51.847578071455088288616446818711
    70.832873937409296415793129090208
    12.524474726415933018910355256388
    37.721878707889458815773572522044
   -39.123034260347914003761194641743
    82.618237465274092122200710770248
    12.591224787366274858850190258322

I expect the following code to be close to zero

    double(subs(polyA,37.721878707889458815773572522044))

But the answer of the matlab is `-2.2191e+11`, other eigenvalues generate other big numbers as well. Why the eigenvalues don't zero the `polyA` while the condition number is low?

[Matt J]

I suggest you attach A in a .mat file, so that we can examine this. But there is no obvious reason why `-2.2191e+11` is to be considered a "big number". If non-eigenvalues generate numbers like 1e+22, then -2.2191e+11 is small by comparison.

[Torsten]

What do you get if you use

e = eig(A)

?

[Mohammad Heidari]

eig(A) is exactly the same numbers I listed as the eigenvalues, but the float point precision is 4 digits.

Answer 1

[Christine Tobler]

The reason for this is that the characteristic polynomial for matrices of even moderate size tends to be ill-conditioned, no matter the condition number of the matrix. For this reason, numerical methods do not use the characteristic polynomial.

For symbolic numbers, the characteristic polynomial works great, but copy-pasting a number such as 37.721878707889458815773572522044 puts you back in the numeric point of view. If you pass in

 s = solve(polyA);
 double(subs(polyA,s(ind)))

this should return zero for every element of s.

To visualize the problem with using the characteristic polynomial, try plotting it:

 x = linspace(-100, 100);
 y = double(subs(polyA, sym(x)));
 plot(x, y);
 ylim([-1e10, 1e10]) % adjust according to the values you see

You should see a bunch of nearly vertical lines crossing the x-axis: a small change in x will result in y being way off.