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1、大连交通大学 2014 届本科生毕业设计（论文）外文翻译1The cyclic nature of the matrix diagonalization method to find a matrix From:http:/.hk/books?hl=zh-CN When A keratosis when can we say that the A diagonalization, referring to seek reversible matrix T so for the angular matrix.1TATMan involved in fundamental theoremTheor
2、em 1:N order matrix A necessary and sufficient condition similar to a 大连交通大学 2014 届本科生毕业设计（论文）外文翻译2diagonal matrix A by n linearly independent eigenvectors, and when A is similar to a diagonal matrix whens main diagonal elements are all eigenvalues.Corollary 1: Matrix A necessary and sufficient cond
3、ition similar to a diagonal matrix of eigenvalues belonging to each of linearly independent eigenvectors A number exactly equal to the multiplicity of the eigenvalues.Theorem 2: If the matrix A of order n has n mutually different eigenvalues (ie eigenvalues of A are single characteristic value), the
4、n A will be similar to a diagonal matrix. The cyclic nature of the matrix diagonalization method to find a matrix1.A basic similarity in through the back diagonal matrixn matrix P through the back called a basic array.010000 001000 000100000010 000001 L L L LLLLLLL L Lit meets on the following prope
5、rties: kP0110n kkIknIn n nPIthe basic matrix obtained through the back of the characteristic polynomial P is:IP10000 01000 0010000001 10000 L L L LLLLLLL L L1nBecause the characteristic polynomial there are n distinct characteristic 10n roots: 22cossin0,1,21 ,kkkiknnnLSo, basically through the back
6、array P is similar to a diagonal matrix.The following feature vector obtained:takeu r11,kn k M大连交通大学 2014 届本科生毕业设计（论文）外文翻译3There（becausePu r010000 001000 000100000010 000001 L L LLLLLLLL L L11kn k M21kk Mk u r1n k）, sois characterized by the rootP corresponding to the eigenvectors.u r11kn k MkAs a m
7、atrix:,T12321 22222 1232122222 12321 11311 12321111111 1 11 1nnnnnnnnn nn nnnnn nn L L L LLLLLLL L LBecause As Determinant，So T reversible, TVandermondeT010,ji ij n then:.1T PT12311n O2. Through the back similar to the diagonal matrixMatrixreferred back to the front through.Q0121101221031230nnnnnncc
8、cccccccccccccc LLLLLLLLLcan find the basic matrix of polynomials out through the back:Q大连交通大学 2014 届本科生毕业设计（论文）外文翻译4Q21 0121n nc Ic Pc PcP LLet： , f x21 0121n ncc xc xcx L1T QT1T21 0121n nc Ic Pc PcP L21111 0121nnc IcT PTcT PTcT PT L 12311,nf f f ff Oso back through the array can be diagonalized. 3.
9、Arbitrary n matrix A can keratosis is a necessary and sufficient condition A similar array of an n-order through the back . Proof:Adequacy: If A is similar to the array through the back so that the presence of reversible matrix C existbut1C ACQ1T QT 12311,nf f f ff Oso1T1C ACT1CTACT 12311,nf f f ff
10、Oie A similar diagonal.Necessity: If A can be diagonalized, which makes the presence of reversible square大连交通大学 2014 届本科生毕业设计（论文）外文翻译5.1C AC12n OWith a polynomial of degree n-1As an equation as f x21 0121n ntt xt xtx Lfollows:,Namely: 11211nnfff L01111 01 11 121 01111nn nn nnnnttttttttt L LL LLL Lde
11、terminant factor in the equation is Determinant,VandermondeT12321 22222 1232122222 12321 11311 123211111111111nnnnnnnnn nn nnnnn nn LLLLLLLLLLLL 010,ji ij n thus the Law known by the unique solution of the equation Let of ordercaramerThe polynomial of degree n-1 is:011,nc ccL, f x21 0121n ncc xc xcx
12、 Ltake matrix, Where P is the basic matrix Q f P21 0121n nc Ic Pc PcP Lthrough the back, thus Q is through the back front, and there is1C AC12n O 12311nf f f ff O大连交通大学 2014 届本科生毕业设计（论文）外文翻译61T QTSo, , that A similar to Q.array through the back.AC1T QT1C111TCQTC大连交通大学 2014 届本科生毕业设计（论文）外文翻译7利用循环矩阵的性质
13、寻找矩阵对角化的方法利用循环矩阵的性质寻找矩阵对角化的方法From:http:/.hk/books?hl=zh-CN当可对角化时,我们说将对角化,即APAA指求可逆矩阵使为对角形矩阵。T1TAT文中涉及的基本定理文中涉及的基本定理定理定理 1:1:阶方阵相似于对角矩阵的充分必要条件是由个线性无关的nAAn特征向量,且当相似于对角矩阵时,的主对角线元素就是的全部特征值。AA推论推论 1:1:方阵相似于对角矩阵的充分必要条件是的属于每个特征值的线AA性无关的特征向量个数正好等于该特征值的重数。定理定理 2:2: 如果阶方阵有个互不相同的特征值（即的特征值都是单特nAnA大连交通大学 2014 届本科生毕业设计（论文）外文翻译8征值）,则必相似于对角矩阵。A利用循环矩阵性质寻找矩阵对角化的方法利用循环矩阵性质寻找矩阵对角化的方法1.1.基本循回阵相似于对角阵基本循回阵相似于对角阵阶矩阵称为基本循回阵。nP010000 001000 000100000010 000001 L L L LLLLLLL L L它满足于如下性质： kP0110n kkIknIn n n
