【正文】 構(gòu)動(dòng)力分析中有重要作用，矩陣迭代法是求矩陣的第一階特征值與特征向量的一種數(shù)值方法 ,但是選取不同的初始向量使結(jié)果可能收斂于不同階的特征值與特征向量，而不一定收斂與第一階 ,陳建兵在《矩陣迭代法求矩陣特征值與特征向量初始向量選取的討論》中 討論了初始向量的選取問題 .特征值理論是線性代數(shù)中的一個(gè)重要的內(nèi)容；當(dāng)方陣階數(shù)很高時(shí)實(shí)際計(jì)算比較繁瑣，趙娜、呂劍峰在《特征值問題的 MATLAB實(shí)踐》中從實(shí)際案例入手，利用 MATLAB軟件討論了求解特征值問題的全過程 .汪慶麗在《用矩陣的初等變換求矩陣的特征值與特征向量》中研究了一種只對(duì)矩陣作適當(dāng)?shù)某醯刃凶儞Q就能求到矩陣的特征值與特征向量的方法，論證其方法的合理性，并闡述此方法的具體求解步驟；岳嶸在《由特征值特征向量去頂矩陣的方法證明及應(yīng)用》中探究了已知 n階對(duì)稱矩陣 A的 k個(gè)互不相等的特征值及 k1個(gè)特征向量計(jì)算出 矩陣 A的計(jì)算方法；張紅玉在《矩陣特征值的理論及應(yīng)用》中討論了通過 n階方陣 A 的特征值得出一系列相關(guān)矩陣的特征值 ,再由特征值與正定矩陣的關(guān)系得出正定矩陣的結(jié)論；劉學(xué)鵬、楊軍在《矩陣的特征值、特征向量和應(yīng)用》一文中討論了矩陣的特征值和特征向量的一些特殊情況，以及在矩陣對(duì)角化方面的應(yīng)用；馮俊艷、馬麗在《討論矩陣的特征值與行列式的關(guān)系》中討論了利用矩陣的特征值解決行列式的問題等等。 在前人研究的基礎(chǔ)上，本文給出了特征值與特征向量的概念及其性質(zhì)，特征值與特征向量性質(zhì)是最基本的內(nèi)容，特征值與特征向量的歸納使得這一工具的使 用更加便利，解決問題的作用更強(qiáng)有力，其應(yīng)用也就更廣泛 .在此基礎(chǔ)上，對(duì)矩陣的特征值與特征向量的計(jì)算進(jìn)行詳盡的闡述和說明 .由于特征值與特征向量的應(yīng)用是多方面的，本文重點(diǎn)介紹了對(duì)特征值與特征向量的應(yīng)用歸納 ,闡述了特征值和特征向量在矩陣運(yùn)算中的作用，以及部分在實(shí)際生活中的應(yīng)用。在例題解析中運(yùn)用一些特征值與特征向量的性質(zhì)和方法，可以使問題更簡(jiǎn)單，運(yùn)算上更方便，是簡(jiǎn)化有關(guān)復(fù)雜問題的一種有效途徑 .本文就是通過大量的例子加以說明運(yùn)用特征值與特征向量的性質(zhì)可以使問題更加清楚，從而使高等代數(shù)中的大量習(xí)題迎刃而解，把特征值與特征 向量在解決實(shí)際問題中的優(yōu)越性表現(xiàn)出來 . 河北師范大學(xué)匯華學(xué)院本科生畢業(yè)論文（設(shè)計(jì)）翻譯文章 27 矩陣的特征值可以確定所發(fā)現(xiàn)的特征多項(xiàng)式的根。多項(xiàng)式的根的顯式代數(shù)公式僅當(dāng)存在比率為 4以下。根據(jù)阿貝爾 魯菲尼定理 5個(gè)或 5個(gè)以上的多項(xiàng)式的根源是沒有一般情況下，明確和準(zhǔn)確的代數(shù)公式。 事實(shí)證明，任何程度的多項(xiàng)式是一些同伴階矩陣的特征多項(xiàng)式。因此， 5個(gè)或更多的順序的矩陣的特征值和特征向量不能獲得通過明確的代數(shù)公式，因此，必須計(jì)算的近似數(shù)值方法 在理論上，可以精確計(jì)算的特征多項(xiàng)式的系數(shù)，因?yàn)樗鼈兪蔷仃囋氐目偤停?算法，可以找到任何所需的精度。然而，任意程度的多項(xiàng)式的所有根這種方法在實(shí)踐中是不可行的，因?yàn)橄禂?shù)將被污染的不可避免的舍入誤差，多項(xiàng)式的根可以是一個(gè)極為敏感的功能（例如由威爾金森的多項(xiàng)式系數(shù)）。 在實(shí)踐中可行，因?yàn)橄禂?shù)將被污染的不可避免的舍入誤差，多項(xiàng)式的根可以是一個(gè)極為敏感的功能（例如由威爾金森的多項(xiàng)式系數(shù)） 直到 QR 算法在 1961年的來臨，高效，精確的方法來計(jì)算任意矩陣的特征值和特征向量。 [與 LU 分解法的查詢結(jié)果在一個(gè)算法中與更好地的 QR 算法的收斂性比。結(jié)合了Householder 變換。對(duì)于大的的厄密共 軛的稀疏矩陣， theLanczos 算法 。是一個(gè)有效的迭代的方法，以計(jì)算特征值和特征向量獲得的一個(gè)例子，在一些其他的可能性。 [編輯 ]計(jì)算特征向量 一旦一個(gè)特征值（精確）的值是已知的，可以找到對(duì)應(yīng)的特征向量，通過尋找特征值方程的非零解，即成為與已知的系數(shù)的線性方程系統(tǒng)。例如，一旦它是已知的，圖 6是矩陣的特征值 我們可以找到它的特征向量，通過求解方程，也就是 ??????????????????? yxyx 636 14 該矩陣方程相當(dāng)于兩個(gè)線性方程組的??? ?? ?? 6y3y6x 6xyx4也就是??? ??? ?? 036 02 yx yx 兩個(gè)方程減少到單一的線性方程 xy 2? .因此，任何載體的形式，任何非零實(shí)數(shù)，是一個(gè)特征值與特征向量相匹配。 上述矩陣 A 有另一個(gè)特征值。類似的計(jì)算表明，對(duì)應(yīng)的特征向量是非零的解決方案，那就是，任何載體的形式，任何非零實(shí)數(shù) b。 某些數(shù)字的方法，計(jì)算的矩陣的特征值也確定一組對(duì)應(yīng)的特征向量作為副產(chǎn)物的計(jì)算。 里昂，線性代數(shù)（第一版）【 M】 .北京：機(jī)械工業(yè)出版社， 2022. The eigenvalues of a matrix can be determined by finding the roots of the characteristic polynomial. Explicit algebraic formulas for the roots of a polynomial exist only if the degree is 4 or less. According to the Abel–Ruffini theorem there is no general, explicit and exact algebraic formula for the roots of a polynomial with degree 5 or more. It turns out that any polynomial with degree is the characteristic polynomial of some panion matrix of order . Therefore, for matrices of order 5 or more, the eigenvalues and eigenvectors cannot be obtained by an explicit algebraic formula, and must therefore be puted by approximate numerical methods. In theory, the coefficients of the characteristic polynomial can be puted exactly, since they 28 are sums of products of matrix elements。 and there are algorithms that can find all the roots of a polynomial of arbitrary degree to any required accuracy.[10] However, this approach is not viable in practice because the coefficients would be contaminated by unavoidable roundoff errors, and the roots of a polynomial can be an extremely sensitive function of the coefficients (as exemplified by Wilkinson39。s polynomial).[10] Efficient, accurate methods to pute eigenvalues and eigenvectors of arbitrary matrices were not known until the advent of the QR algorithm in 1961. [10] Combining the Householder transformation with the LU deposition results in an algorithm with better convergence than the QR algorithm.[11] For large Hermitian sparse matrices, theLanczos algorithm is one example of an efficient iterative method to pute eigenvalues and eigenvectors, among several other possibilities.[10] [edit] Computing the eigenvectors Once the (exact) value of an eigenvalue is known, the corresponding eigenvectors can be found by finding nonzero solutions of the eigenvalue equation, that bees a system of linear equations with known coefficients. For example, once it is known that 6 is an eigenvalue of the matrix ??????? 36 14A we can find its eigenvectors by solving the equation V6AV? ， that is ??????????????????? yxyx 636 14 This matrix equation is equivalent to two linear equations??? ?? ?? 6y3y6x 6xyx4that is??? ??? ?? 036 02 yx yx Both equations reduce to the single linear equation xy 2? . Therefore, any vector of the form ? ?39。2, aa ,for any nonzero real number a, is an eigenvector of A with eigenvalue 6?? . The matrix A above has another eigenvalue 1?? . A similar calculation shows that the corresponding eigenvectors are the nonzero solutions of 03 ??yx , that is, any vector of the form? ?39。3, bb? , for any nonzero real number . Some numeric methods that pute the eigenvalues of a matrix also determine a set of corresponding eigenvectors as a byproduct of the putation. ag an employment tribunal clai Emloyment tribunals sort out disagreements between employers and employees. You may need to make a claim to an employment tribunal if: you don39。t agree with the disciplinary action your employer has taken against you your employer dismisses you and you think that you have been dismissed unfairly. For more informu, take advice from one of the anisations listed under Further help. Employment tribunals are less formal than some other courts, but it is still a legal process and you will need to give evidence under an oath or affirmation. Most people find making a claim to an employment tribunal challenging. If you are thinking about making a claim to an employment tribunal, you should get help straight away from one of the anisations listed under Further help. ation about dismissal and unfair dismissal, see Dismissal. You can make a claim to an employment tribunal, even if you haven39。t appealed against the disciplinary action your employer has taken against you. However, if you win your case, the tribunal may reduce any pensat