矩阵对角化相关推导

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前提条件：

有一个 $n$ 阶可逆矩阵 $A$ 。

A = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 \dots \dots ⋱ \dots a 1 n a 2 n ⋮ a n n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟

$A=\begin{pmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\\end{pmatrix}$
对于行列式为：

| A | = ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 \dots \dots ⋱ \dots a 1 n a 2 n ⋮ a n n ∣ ∣ ∣ ∣ ∣ ∣ ∣

$|A|=\left|\begin{matrix}a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\\end{matrix}\right|$
则其伴随矩阵为：

A * = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ A 11 A 12 ⋮ A 1 n A 21 A 22 ⋮ A 2 n \dots \dots ⋱ \dots A n 1 A n 2 ⋮ A n n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟

$A^*=\begin{pmatrix}A_{11}&A_{21}&\cdots&A_{n1}\\ A_{12}&A_{22}&\cdots&A_{n2}\\ \vdots&\vdots&\ddots&\vdots\\ A_{1n}&A_{2n}&\cdots&A_{nn}\\\end{pmatrix}$
则逆矩阵为：

A - 1 = A * | A |

$A^{-1}=\dfrac{A^*}{|A|}$
为什么

A−1A−1 $A^{-1}$ 会是这样

验证：
前提
行列式按照某一行展开：

| A | = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 11 a 21 ⋮ a i 1 ⋮ a n 1 a 12 a 22 ⋮ a i 2 ⋮ a n 2 \dots \dots \dots \dots a 1 n a 2 n ⋮ a i n ⋮ a n n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = a i 1 A i 1 + a i 2 A i 2 + \dots + a i n A i n

$|A|=\left|\begin{matrix}a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\space&\vdots\\ \color{red}{a_{i1}}&\color{red}{a_{i2}}&\cdots&\color{red}{a_{in}}\\ \vdots&\vdots&\space&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\\end{matrix}\right|=\color{red}{a_{i1}}A_{i1}+\color{red}{a_{i2}}A_{i2}+\cdots+\color{red}{a_{in}}A_{in}$

所以如果是：

a j 1 A i 1 + a j 2 A i 2 + \dots + a j n A i n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 11 a 21 ⋮ a j 1 ⋮ a j 1 ⋮ a n 1 a 12 a 22 ⋮ a j 2 ⋮ a j 2 ⋮ a n 2 \dots \dots \dots \dots \dots a 1 n a 2 n ⋮ a j n ⋮ a j n ⋮ a n n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0

$\color{blue}{a_{j1}}A_{i1}+\color{blue}{a_{j2}}A_{i2}+\cdots+\color{blue}{a_{jn}}A_{in}=\left|\begin{matrix}a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\space&\vdots\\ \color{blue}{a_{j1}}&\color{blue}{a_{j2}}&\cdots&\color{blue}{a_{jn}}\\ \vdots&\vdots&\space&\vdots\\ \color{blue}{a_{j1}}&\color{blue}{a_{j2}}&\cdots&\color{blue}{a_{jn}}\\ \vdots&\vdots&\space&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\\end{matrix}\right|=0$
为什么

A−1=A∗|A|A−1=A∗|A| $A^{-1}=\dfrac{A^*}{|A|}$
因为：

A - 1 A = A * | A | A = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ A 11 | A | A 12 | A | ⋮ A 1 n | A | A 21 | A | A 22 | A | ⋮ A 2 n | A | \dots \dots ⋱ \dots A n 1 | A | A n 2 | A | ⋮ A n n | A | ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 \dots \dots ⋱ \dots a 1 n a 2 n ⋮ a n n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ a 11 A 11 + \dots + a n 1 A n 1 | A | a 11 A 12 + \dots + a n 1 A n 2 | A | ⋮ a 11 A 1 n + \dots + a n 1 A n n | A | a 12 A 11 + \dots + a n 2 A n 1 | A | a 12 A 11 + \dots + a n 2 A n 1 | A | ⋮ a 12 A 1 n + \dots + a n 2 A n n | A | \dots \dots ⋱ \dots a 1 n A 11 + \dots + a n n A n 1 | A | a 1 n A 11 + \dots + a n n A n 1 | A | ⋮ a 1 n A 1 n + \dots + a n n A n n | A | ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ | A | | A | 0 | A | ⋮ 0 | A | 0 | A | | A | | A | ⋮ 0 | A | \dots \dots ⋱ \dots 0 | A | 0 | A | ⋮ | A | | A | ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ 10 ⋮ 0 01 ⋮ 0 \dots \dots ⋱ \dots 00 ⋮ 1 ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ = E

$\begin{aligned} A^{-1}A&=\dfrac{A^*}{|A|}A=\begin{pmatrix}\dfrac{A_{11}}{|A|}&\dfrac{A_{21}}{|A|}&\cdots&\dfrac{A_{n1}}{|A|}\\ \dfrac{A_{12}}{|A|}&\dfrac{A_{22}}{|A|}&\cdots&\dfrac{A_{n2}}{|A|}\\ \vdots&\vdots&\ddots&\vdots\\ \dfrac{A_{1n}}{|A|}&\dfrac{A_{2n}}{|A|}&\cdots&\dfrac{A_{nn}}{|A|}\\\end{pmatrix}\begin{pmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\\end{pmatrix}\\ \\ &=\begin{pmatrix}\dfrac{\color{red}{a_{11}}A_{11}+\cdots+\color{red}{a_{n1}}A_{n1}}{|A|}&\dfrac{\color{blue}{a_{12}}A_{11}+\cdots+\color{blue}{a_{n2}}A_{n1}}{|A|}&\cdots&\dfrac{\color{blue}{a_{1n}}A_{11}+\cdots+\color{blue}{a_{nn}}A_{n1}}{|A|}\\\dfrac{\color{blue}{a_{11}}A_{12}+\cdots+\color{blue}{a_{n1}}A_{n2}}{|A|}&\dfrac{\color{red}{a_{12}}A_{11}+\cdots+\color{red}{a_{n2}}A_{n1}}{|A|}&\cdots&\dfrac{\color{blue}{a_{1n}}A_{11}+\cdots+\color{blue}{a_{nn}}A_{n1}}{|A|}\\ \vdots&\vdots&\ddots&\vdots\\\dfrac{\color{blue}{a_{11}}A_{1n}+\cdots+\color{blue}{a_{n1}}A_{nn}}{|A|}&\dfrac{\color{blue}{a_{12}}A_{1n}+\cdots+\color{blue}{a_{n2}}A_{nn}}{|A|}&\cdots&\dfrac{\color{red}{a_{1n}}A_{1n}+\cdots+\color{red}{a_{nn}}A_{nn}}{|A|}\\\end{pmatrix}\\ &=\begin{pmatrix}\dfrac{|A|}{|A|}&\dfrac{0}{|A|}&\cdots&\dfrac{0}{|A|}\\ \dfrac{0}{|A|}&\dfrac{|A|}{|A|}&\cdots&\dfrac{0}{|A|}\\ \vdots&\vdots&\ddots&\vdots\\ \dfrac{0}{|A|}&\dfrac{0}{|A|}&\cdots&\dfrac{|A|}{|A|}\\\end{pmatrix}=\begin{pmatrix}1&0&\cdots&0\\ 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1\\\end{pmatrix}=E \end{aligned}$

A - 1 A = A * | A | A = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ A 11 | A | A 12 | A | ⋮ A 1 n | A | A 21 | A | A 22 | A | ⋮ A 2 n | A | \dots \dots ⋱ \dots A n 1 | A | A n 2 | A | ⋮ A n n | A | ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 \dots \dots ⋱ \dots a 1 n a 2 n ⋮ a n n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ β T 1 β T 2 ⋮ β T n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ (α 1, α 2, \dots, α n)

显然：

β T i α j = {0, i \neq j 1, i = j

$\beta_i^T\alpha_j=\left\{\begin{aligned}0,i\neq j\\ 1,i=j\end{aligned}\right.$

在矩阵的对角化中，
假设矩阵 $A$ 的特征值为 $\lambda_1,\lambda_2,\cdots,\lambda_n$ ,对应的特征向量为 $\alpha_1,\alpha_2,\cdots,\alpha_n$
为什么由特征向量构成的矩阵

P = (α 1, α 2, \dots, α n)

$P=(\alpha_1,\alpha_2,\cdots,\alpha_n)$
若要矩阵

PP $P$ 可逆，则

α_{1}, α_{2}, \dots, α_{n}

$\alpha_1,\alpha_2,\cdots,\alpha_n$ 线性无关
则此时：

P - 1 A P = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ β T 1 β T 2 ⋮ β T n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ A (α 1, α 2, \dots, α n) = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ β T 1 β T 2 ⋮ β T n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ (A α 1, A α 2, \dots, A α n) = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ β T 1 β T 2 ⋮ β T n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ (λ 1 α 1, λ 2 α 2, \dots, λ n α n) = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ β T 1 λ 1 α 1 β T 2 λ 1 α 1 ⋮ β T n λ 1 α 1 β T 1 λ 2 α 2 β T 2 λ 2 α 2 ⋮ β T n λ 2 α 2 \dots \dots ⋱ \dots β T 1 λ n α n β T 2 λ n α n ⋮ β T n λ n α n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ β T 1 λ 1 α 1 0 ⋮ 0 0 β T 2 λ 2 α 2 ⋮ 0 \dots \dots ⋱ \dots 00 ⋮ β T n λ n α n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ λ 1 0 ⋮ 0 0 λ 2 ⋮ 0 \dots \dots ⋱ \dots 00 ⋮ λ n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟

$\begin{aligned} P^{-1}AP&=\begin{pmatrix}\beta_1^T\\\beta_2^T\\\vdots\\\beta_n^T\end{pmatrix}A(\alpha_1,\alpha_2,\cdots,\alpha_n)\\ \\ &=\begin{pmatrix}\beta_1^T\\\beta_2^T\\\vdots\\\beta_n^T\end{pmatrix}(A\alpha_1,A\alpha_2,\cdots,A\alpha_n)=\begin{pmatrix}\beta_1^T\\\beta_2^T\\\vdots\\\beta_n^T\end{pmatrix}(\lambda_1\alpha_1,\lambda_2\alpha_2,\cdots,\lambda_n\alpha_n)\\ &=\begin{pmatrix}\beta_1^T\lambda_1\alpha_1&\beta_1^T\lambda_2\alpha_2&\cdots&\beta_1^T\lambda_n\alpha_n\\ \beta_2^T\lambda_1\alpha_1&\beta_2^T\lambda_2\alpha_2&\cdots&\beta_2^T\lambda_n\alpha_n\\ \vdots&\vdots&\ddots&\vdots\\ \beta_n^T\lambda_1\alpha_1&\beta_n^T\lambda_2\alpha_2&\cdots&\beta_n^T\lambda_n\alpha_n\\ \end{pmatrix}\\ \\ &=\begin{pmatrix}\beta_1^T\lambda_1\alpha_1&0&\cdots&0\\ 0&\beta_2^T\lambda_2\alpha_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\beta_n^T\lambda_n\alpha_n\\ \end{pmatrix}\\ \\ &=\begin{pmatrix}\lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n\\ \end{pmatrix}\\ \end{aligned}$