Let .
Consider , that is uk+1= Auk , where .
Because the characteristic polynomial of A is, the eigenvalues of A areand, and the corresponding eigenvectors are and . Of course {x1, x2} are linearly independent and generate uk.
Write u0=c1x1+c2x2, that is, and it’s easy to solve the coefficients and . Hence
u1=c1Ax1+c2Ax2= c1λ1x1+c2λ2x2 ,
… ,
Thus
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