svd prescision is very bad.
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it appears to be that when i use SVD i loose prescision how can i avoid loosing prescision and use svd function?
[U,S,V]=svd(T);
T=U*S*V'
the first T Matrix and the second are not the same.
here a comparation of the matrix before svd and after:
>> T
T =
-0.4609 + 0.4970i 0.0023 + 0.0267i -0.0267 + 0.0028i
0.0023 + 0.0270i -0.5192 - 0.4982i -0.0023 - 0.0267i
-0.0267 + 0.0028i -0.0023 - 0.0270i -0.4609 + 0.4970i
>> [U,S,V]=svd(T); >> Tsvd=U*S*V'
Tsvd =
-0.4609 + 0.4970i 0.0023 + 0.0267i -0.0267 + 0.0028i
0.0023 + 0.0270i -0.5192 - 0.4982i -0.0023 - 0.0267i
-0.0267 + 0.0028i -0.0023 - 0.0270i -0.4609 + 0.4970i
>> difference=T-Tsvd
difference =
1.0e-15 *
-0.0555 - 0.1110i 0.0247 - 0.0312i -0.4025 + 0.3092i
-0.0278 - 0.0173i 0.0000 - 0.3331i -0.0494 + 0.0555i
-0.0486 + 0.0867i 0.0694 + 0.1076i 0.0000 + 0.0555i
4 个评论
Roger Stafford
2014-10-14
Kobi, that is just expected round-off error out at the fifteenth decimal place. You can't expect any better precision than that using double precision floating point numbers. After all, these numbers have only 53 bits in their significands. Your description of "very bad" is quite unfair.
采纳的回答
Andreas Goser
2014-10-14
Please let us know how familiar your are with numerical mathematics. The effect you see here is to be expected, but I do not want to come across as too blunt just pointing you to
eps
I could find a document that describes a bit about the why.
3 个评论
Oleg Komarov
2014-10-14
编辑:Oleg Komarov
2014-10-14
Where do you take 25 digits from?
>> fprintf('%.20f\n',pi)
3.14159265358979310000
>> fprintf('%.20f\n',eps(pi))
0.00000000000000044409
>> fprintf('%.20f\n',pi+eps(pi))
3.14159265358979360000
更多回答(1 个)
Roger Stafford
2014-10-14
编辑:Roger Stafford
2014-10-14
You cannot expect them to be exactly the same because of rounding errors. Have you compared them using "format long" to see how significant the differences are?
If you are still unsatisfied, please give a representative sample of what you have observed.
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