矩阵对角化

先决条件：
特征值和特征向量

设 A 和 B 是两个 n 阶矩阵。如果存在可逆矩阵 P，则可以认为 B 与 A 相似，使得

$B=P^{-1} A P$

这称为矩阵相似变换。

矩阵的对角化定义为将任意矩阵 A 化简为对角形式 D 的过程。根据相似变换，如果矩阵 A 与 D 相关，则

$D = P ^{-1} A P$

并且矩阵A通过另一个矩阵P简化为对角矩阵D。 （P≡模态矩阵）

Modal matrix: It is a (n x n) matrix that consists of eigen-vectors. It is generally used in the process of diagonalization and similarity transformation.

简而言之，它是将方阵转换为一种特殊类型的矩阵的过程，称为对角矩阵。

涉及的步骤

Step 1 - Initialize the diagonal matrix D as:

$D=\left[\begin{array}{ccc} \lambda_{1} & 0 & 0 \\ 0 & \lambda_{2} & 0 \\ 0 & 0 & \lambda_{3} \end{array}\right]$

where λ1, λ2, λ3 -> eigen values

Step 2 - Find the eigen values using the equation given below.

$\operatorname{det}(A-\lambda I)=0$

where, 
A -> given 3x3 square matrix.
I -> identity matrix of size 3x3.
λ -> eigen value.

Step 3 - Compute the corresponding eigen vectors using the equation given below.

$\begin{array}{l} A t, \lambda=i \\ {[A-\lambda I] X_{i}=0} \end{array}$

where,
λi -> eigen value.
Xi -> corresponding eigen vector.

Step 4 - Create the modal matrix P.

$P=\left[X_{0} X_{1} . . X_{n}\right]$

Here, all the eigen vectors till Xi are filled column wise in matrix P. 

Step 5 - Find P-1 and then use equation given below to find diagonal matrix D.

$D=P^{-1} A P$

示例问题

问题陈述：假设一个 3×3 方阵 A 具有以下值：

$A=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 1 & 2 & 1 \\ 2 & 2 & 3 \end{array}\right]$

使用矩阵的对角化找到 A 的对角矩阵 D。 [ D = P ^-1 AP]

逐步解决方案：

Step 1 - Initializing D as:

$D=\left[\begin{array}{ccc} \lambda_{1} & 0 & 0 \\ 0 & \lambda_{2} & 0 \\ 0 & 0 & \lambda_{3} \end{array}\right]$

Step 2 - Find the eigen values. (or possible values of λ)

$\operatorname{det}(A-\lambda I)=0$

$\begin{array}{l} \Longrightarrow \operatorname{det}(A-\lambda I)=\operatorname{det}\left(\left[\begin{array}{ccc} 1-\lambda & 0 & -1 \\ 1 & 2-\lambda & 1 \\ 2 & 2 & 3-\lambda \end{array}\right]\right)=0 \\ \Longrightarrow\left(\lambda^{3}-6 \lambda^{2}+11 \lambda-6\right)=0 & \\ \Longrightarrow(\lambda-1)(\lambda-2)(\lambda-3)=0 \\ \Longrightarrow & \lambda=1,2,3 \end{array}$

Step 3 - Find the eigen vectors X1, X2, X3 corresponding to the eigen values λ = 1,2,3. 

$At $\lambda=1$ A - $(1) I$ $X_{1}=0$ $\Longrightarrow\left[\begin{array}{ccc}1-1 & 0 & -1 \\ 1 & 2-1 & 1 \\ 2 & 2 & 3-1\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ $\Longrightarrow\left[\begin{array}{ccc}0 & 0 & -1 \\ 1 & 1 & 1 \\ 2 & 2 & 2\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ On solving, we get the following equations: $x_{3}=0\left(x_{1}\right)$ $x_{1}+x_{2}=0 \Longrightarrow x_{2}=-x_{1}$ $\therefore X_{1}=\left[\begin{array}{c}x_{1} \\ -x_{1} \\ 0\left(x_{1}\right)\end{array}\right]$ $\Longrightarrow X_{1}=\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]$ Similarly, for $\lambda=2$ $X_{2}=\left[\begin{array}{c}-2 \\ 1 \\ 2\end{array}\right]$ and for $\lambda=3$ $X_{3}=\left[\begin{array}{c}1 \\ -1 \\ -2\end{array}\right]$$

$Similarly, \\ for\ \lambda = 2 \\ X_2 = \begin{bmatrix} -2\\1\\2 \end{bmatrix} \\ and \\ for\ \lambda = 3 \\ X_3 = \begin{bmatrix} 1\\-1\\-2 \end{bmatrix}$

Step 5 - Creation of modal matrix P. (here, X1, X2, X3 are column vectors)

$P=\left[X_{1} X_{2} X_{3}\right]=\left[\begin{array}{ccc} 1 & -2 & 1 \\ -1 & 1 & -1 \\ 0 & 2 & -2 \end{array}\right]$

Step 6 - Finding P-1 and then putting values in diagonalization of a matrix equation. [D = P-1AP]

We do Step 6 to find out which eigen value will replace λ₁, λ₂ and λ₃ in the initial diagonal matrix created in Step 1.

$\begin{array}{l} \begin{array}{l} \quad P=\left[\begin{array}{ccc} 1 & -2 & 1 \\ -1 & 1 & -1 \\ 0 & 2 & -2 \end{array}\right] \\ \operatorname{det}(P)=(1)[(-2)(1)-(-1)(2)]-(-2)[(-2)(-1)-(0)(-1)]+(1)[(2)(-1)- \\ (0)(1)] \\ =[0+(4)+(-2)] \\ =2 \end{array}\\ \text { Since } \operatorname{det}(P) \neq 0 \Longrightarrow \text { Matrix } P \text { is invertible. } \end{array}$

Reference articles: Determinant of a matrix & Inverse of a matrix

我们知道

$P^{-1}=\frac{\operatorname{adj}(P)}{\operatorname{det}(P)}$

在求解时，我们得到

$P^{-1}=\frac{1}{2}\left[\begin{array}{ccc} 0 & -2 & 1 \\ -2 & -2 & 0 \\ -2 & -2 & -1 \end{array}\right]$

代入矩阵方程的对角化，我们得到

$\begin{array}{l} \quad D=P^{-1} A P \\ D=\frac{1}{2}\left[\begin{array}{ccc} 0 & -2 & 1 \\ -2 & -2 & 0 \\ -2 & -2 & -1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 1 & 2 & 1 \\ 2 & 2 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & -2 & 1 \\ -1 & 1 & -1 \\ 0 & 2 & -2 \end{array}\right] \\ D=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right] \end{array}$