Eigenvalue doesn't zero the characteristic polynomial

2018 8 14
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更新时间：2018 8 14
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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?

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Matt J
Matt J 2018-8-14
编辑：Matt J 2018-8-14
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.
Mohammad Heidari
Mohammad Heidari 2018-8-14
eig(A) is exactly the same numbers I listed as the eigenvalues, but the float point precision is 4 digits.

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 采纳的回答

Christine Tobler
Christine Tobler 2018-8-14

2 个投票

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.

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