矩陣對角化相關推導

前提條件：

有一個n$n$ 階可逆矩陣A$A$ 。

A=⎛⎝⎜⎜⎜⎜⎜a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎞⎠⎟⎟⎟⎟⎟$$A=\left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right)$$

對於行列式爲：

|A|=∣∣∣∣∣∣∣a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann∣∣∣∣∣∣∣$$|A|=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|$$

則其伴隨矩陣爲：

A∗=⎛⎝⎜⎜⎜⎜⎜A11A12⋮A1nA21A22⋮A2n⋯⋯⋱⋯An1An2⋮Ann⎞⎠⎟⎟⎟⎟⎟$$A^{\ast }=\left(\begin{array}{cccc}{A}_{11}& A_{21}& \cdots & A_{n1}\\ A_{12}& A_{22}& \cdots & A_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ A_{1n}& A_{2n}& \cdots & A_{nn}\end{array}\right)$$

則逆矩陣爲：

A−1=A∗|A|$$A^{-1}={\displaystyle \frac{A^{\ast }}{|A|}}$$

爲什麼A−1$A^{-1}$ 會是這樣

驗證：
前提
行列式按照某一行展開：

|A|=∣∣∣∣∣∣∣∣∣∣a11a21⋮ai1⋮an1a12a22⋮ai2⋮an2⋯⋯ ⋯ ⋯a1na2n⋮ain⋮ann∣∣∣∣∣∣∣∣∣∣=ai1Ai1+ai2Ai2+⋯+ainAin$$|A|=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{i1}& a_{i2}& \cdots & a_{in}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|=a_{i1}{A}_{i1}+a_{i2}{A}_{i2}+\cdots +a_{in}{A}_{in}$$

所以如果是：

aj1Ai1+aj2Ai2+⋯+ajnAin=∣∣∣∣∣∣∣∣∣∣∣∣∣∣a11a21⋮aj1⋮aj1⋮an1a12a22⋮aj2⋮aj2⋮an2⋯⋯ ⋯ ⋯ ⋯a1na2n⋮ajn⋮ajn⋮ann∣∣∣∣∣∣∣∣∣∣∣∣∣∣=0$$a_{j1}{A}_{i1}+a_{j2}{A}_{i2}+\cdots +a_{jn}{A}_{in}=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{j1}& a_{j2}& \cdots & a_{jn}\\ ⋮& ⋮& & ⋮\\ a_{j1}& a_{j2}& \cdots & a_{jn}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|=0$$

爲什麼A−1=A∗|A|$A^{-1}={\displaystyle \frac{A^{\ast }}{|A|}}$
因爲：

A−1A=A∗|A|A=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜A11|A|A12|A|⋮A1n|A|A21|A|A22|A|⋮A2n|A|⋯⋯⋱⋯An1|A|An2|A|⋮Ann|A|⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎛⎝⎜⎜⎜⎜⎜a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎞⎠⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜a11A11+⋯+an1An1|A|a11A12+⋯+an1An2|A|⋮a11A1n+⋯+an1Ann|A|a12A11+⋯+an2An1|A|a12A11+⋯+an2An1|A|⋮a12A1n+⋯+an2Ann|A|⋯⋯⋱⋯a1nA11+⋯+annAn1|A|a1nA11+⋯+annAn1|A|⋮a1nA1n+⋯+annAnn|A|⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜|A||A|0|A|⋮0|A|0|A||A||A|⋮0|A|⋯⋯⋱⋯0|A|0|A|⋮|A||A|⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜10⋮001⋮0⋯⋯⋱⋯00⋮1⎞⎠⎟⎟⎟⎟⎟=E$$\begin{array}{rl}{A}^{-1}A& ={\displaystyle \frac{A^{\ast }}{|A|}}A=\left(\begin{array}{cccc}{\displaystyle \frac{A_{11}}{|A|}}& {\displaystyle \frac{A_{21}}{|A|}}& \cdots & {\displaystyle \frac{A_{n1}}{|A|}}\\ {\displaystyle \frac{A_{12}}{|A|}}& {\displaystyle \frac{A_{22}}{|A|}}& \cdots & {\displaystyle \frac{A_{n2}}{|A|}}\\ ⋮& ⋮& \ddots & ⋮\\ {\displaystyle \frac{A_{1n}}{|A|}}& {\displaystyle \frac{A_{2n}}{|A|}}& \cdots & {\displaystyle \frac{A_{nn}}{|A|}}\end{array}\right)\left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right)\\ \\ & =\left(\begin{array}{cccc}{\displaystyle \frac{a_{11}{A}_{11}+\cdots +a_{n1}{A}_{n1}}{|A|}}& {\displaystyle \frac{a_{12}{A}_{11}+\cdots +a_{n2}{A}_{n1}}{|A|}}& \cdots & {\displaystyle \frac{a_{1n}{A}_{11}+\cdots +a_{nn}{A}_{n1}}{|A|}}\\ {\displaystyle \frac{a_{11}{A}_{12}+\cdots +a_{n1}{A}_{n2}}{|A|}}& {\displaystyle \frac{a_{12}{A}_{11}+\cdots +a_{n2}{A}_{n1}}{|A|}}& \cdots & {\displaystyle \frac{a_{1n}{A}_{11}+\cdots +a_{nn}{A}_{n1}}{|A|}}\\ ⋮& ⋮& \ddots & ⋮\\ {\displaystyle \frac{a_{11}{A}_{1n}+\cdots +a_{n1}{A}_{nn}}{|A|}}& {\displaystyle \frac{a_{12}{A}_{1n}+\cdots +a_{n2}{A}_{nn}}{|A|}}& \cdots & {\displaystyle \frac{a_{1n}{A}_{1n}+\cdots +a_{nn}{A}_{nn}}{|A|}}\end{array}\right)\\ & =\left(\begin{array}{cccc}{\displaystyle \frac{|A|}{|A|}}& {\displaystyle \frac{0}{|A|}}& \cdots & {\displaystyle \frac{0}{|A|}}\\ {\displaystyle \frac{0}{|A|}}& {\displaystyle \frac{|A|}{|A|}}& \cdots & {\displaystyle \frac{0}{|A|}}\\ ⋮& ⋮& \ddots & ⋮\\ {\displaystyle \frac{0}{|A|}}& {\displaystyle \frac{0}{|A|}}& \cdots & {\displaystyle \frac{|A|}{|A|}}\end{array}\right)=\left(\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 1\end{array}\right)=E\end{array}$$

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A−1A=A∗|A|A=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜A11|A|A12|A|⋮A1n|A|A21|A|A22|A|⋮A2n|A|⋯⋯⋱⋯An1|A|An2|A|⋮Ann|A|⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎛⎝⎜⎜⎜⎜⎜a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎞⎠⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜βT1βT2⋮βTn⎞⎠⎟⎟⎟⎟⎟(α1,α2,⋯,αn)$$\begin{array}{rl}{A}^{-1}A& ={\displaystyle \frac{A^{\ast }}{|A|}}A=\left(\begin{array}{cccc}{\displaystyle \frac{A_{11}}{|A|}}& {\displaystyle \frac{A_{21}}{|A|}}& \cdots & {\displaystyle \frac{A_{n1}}{|A|}}\\ {\displaystyle \frac{A_{12}}{|A|}}& {\displaystyle \frac{A_{22}}{|A|}}& \cdots & {\displaystyle \frac{A_{n2}}{|A|}}\\ ⋮& ⋮& \ddots & ⋮\\ {\displaystyle \frac{A_{1n}}{|A|}}& {\displaystyle \frac{A_{2n}}{|A|}}& \cdots & {\displaystyle \frac{A_{nn}}{|A|}}\end{array}\right)\left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right)\\ & =\left(\begin{array}{c}{\beta }_1^T\\ {\beta }_2^T\\ ⋮\\ {\beta }_n^T\end{array}\right)({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)\end{array}$$

顯然：

βTiαj={0,i≠j1,i=j$${\beta }_i^T{\alpha }_j=\{\begin{array}{r}0,i\ne j\\ 1,i=j\end{array}$$

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在矩陣的對角化中，
假設矩陣A$A$ 的特徵值爲λ1,λ2,⋯,λn${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$ ,對應的特徵向量爲α1,α2,⋯,αn${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$
爲什麼由特徵向量構成的矩陣

P=(α1,α2,⋯,αn)$$P=({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)$$

若要矩陣P$P$ 可逆，則α1,α2,⋯,αn${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$ 線性無關
則此時：

P−1AP=⎛⎝⎜⎜⎜⎜⎜βT1βT2⋮βTn⎞⎠⎟⎟⎟⎟⎟A(α1,α2,⋯,αn)=⎛⎝⎜⎜⎜⎜⎜βT1βT2⋮βTn⎞⎠⎟⎟⎟⎟⎟(Aα1,Aα2,⋯,Aαn)=⎛⎝⎜⎜⎜⎜⎜βT1βT2⋮βTn⎞⎠⎟⎟⎟⎟⎟(λ1α1,λ2α2,⋯,λnαn)=⎛⎝⎜⎜⎜⎜⎜βT1λ1α1βT2λ1α1⋮βTnλ1α1βT1λ2α2βT2λ2α2⋮βTnλ2α2⋯⋯⋱⋯βT1λnαnβT2λnαn⋮βTnλnαn⎞⎠⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜βT1λ1α10⋮00βT2λ2α2⋮0⋯⋯⋱⋯00⋮βTnλnαn⎞⎠⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜λ10⋮00λ2⋮0⋯⋯⋱⋯00⋮λn⎞⎠⎟⎟⎟⎟⎟$$\begin{array}{rl}{P}^{-1}AP& =\left(\begin{array}{c}{\beta }_1^T\\ {\beta }_2^T\\ ⋮\\ {\beta }_n^T\end{array}\right)A({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)\\ \\ & =\left(\begin{array}{c}{\beta }_1^T\\ {\beta }_2^T\\ ⋮\\ {\beta }_n^T\end{array}\right)(A{\alpha }_1,A{\alpha }_2,\cdots ,A{\alpha }_n)=\left(\begin{array}{c}{\beta }_1^T\\ {\beta }_2^T\\ ⋮\\ {\beta }_n^T\end{array}\right)({\lambda }_1{\alpha }_1,{\lambda }_2{\alpha }_2,\cdots ,{\lambda }_n{\alpha }_n)\\ & =\left(\begin{array}{cccc}{\beta }_1^T{\lambda }_1{\alpha }_1& {\beta }_1^T{\lambda }_2{\alpha }_2& \cdots & {\beta }_1^T{\lambda }_n{\alpha }_n\\ {\beta }_2^T{\lambda }_1{\alpha }_1& {\beta }_2^T{\lambda }_2{\alpha }_2& \cdots & {\beta }_2^T{\lambda }_n{\alpha }_n\\ ⋮& ⋮& \ddots & ⋮\\ {\beta }_n^T{\lambda }_1{\alpha }_1& {\beta }_n^T{\lambda }_2{\alpha }_2& \cdots & {\beta }_n^T{\lambda }_n{\alpha }_n\end{array}\right)\\ \\ & =\left(\begin{array}{cccc}{\beta }_1^T{\lambda }_1{\alpha }_1& 0& \cdots & 0\\ 0& {\beta }_2^T{\lambda }_2{\alpha }_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\beta }_n^T{\lambda }_n{\alpha }_n\end{array}\right)\\ \\ & =\left(\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_n\end{array}\right)\end{array}$$