s********y
发帖数: 660 | 1 A= 2 -1 0 0 ... 0 0
-1 2 -1 0 ... 0 0
0 -1 2 -1 ... 0 0
...
0 0 0 0 ... 2 -1
0 0 0 0 ... -1 2
A就是个tridiagonal matrix. 对角线上元素[-1 2 -1] | x******r
发帖数: 367 | 2 Let x=(sin(a),sin(2a),...,sin(na))T
If a=j*pi/(n+1),pi-3.1415926...
then x is an eigenvector of A corresponding to the eigenvalue 2-2cos(a).
The proof can be found on page 621
of Numerical analysis: Mathematics of Scientific Computing(third Edition).
The authors are David Kincaid & Ward Cheney.
Hope that it will be helpful.
【在 s********y 的大作中提到】
: A= 2 -1 0 0 ... 0 0
: -1 2 -1 0 ... 0 0
: 0 -1 2 -1 ... 0 0
: ...
: 0 0 0 0 ... 2 -1
: 0 0 0 0 ... -1 2
: A就是个tridiagonal matrix. 对角线上元素[-1 2 -1]
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