フィルターのクリア

Square root of a matrix

4 ビュー (過去 30 日間)
Qian Feng
Qian Feng 2017 年 2 月 21 日
編集済み: Matt J 2017 年 2 月 21 日
I encountered a problem when I try to compute the square root of a positive definite matrix
syms x real
mi = [1; exp(x); exp(2*x); exp(3*x); exp(-x)];
F = vpa(simplify(int(mi*mi',x,-10,0), 'Steps', 100)); Fs = sqrtm(F);
The problem here is Fs should be a symmetric matrix but Fs- Fs' here is not a zero matrix.
I have tried to use symbolic calculation instead of vpa but it seems that it requires a very long time to compute.
Is there any way that Fs can be computed without destroying its symmetric structure ?

採用された回答

Matt J
Matt J 2017 年 2 月 21 日
編集済み: Matt J 2017 年 2 月 21 日
You can post-correct the asymmetry as follows,
Fs=(Fs+Fs.')/2

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeLinear Algebra についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by