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Matrix Diagonalization

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  • Last Updated : 27 Apr, 2022

Prerequisite: Eigenvalues  and eigenvectors

Let A and B be two matrices of order n. B can be considered similar to A if there exists an invertible matrix P such that B=P^{-1} A P  This is known as Matrix Similarity Transformation.

Diagonalization of a matrix is defined as the process of reducing any matrix A into its diagonal form D. As per the similarity transformation, if the matrix A is related to D, then

D = P ^{-1} A P  and the matrix A is reduced to the diagonal matrix D through another matrix P. Where P is a modal matrix) 

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.

In simpler words, it is the process of taking a square matrix and converting it into a special type of matrix called a diagonal matrix.

Steps Involved:

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 3×3 square matrix. I -> identity matrix of size 3×3. 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 eigenvectors till Xi have filled column-wise in matrix P. 

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

D=P^{-1} A P

Example Problem:

Problem Statement: Assume a 3×3 square matrix A having the following values:

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

Find the diagonal matrix D of A using the diagonalization of the matrix. [ D = P-1AP ]

Solution:
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 eigenvalue 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}  

We know that P^{-1}=\frac{\operatorname{adj}(P)}{\operatorname{det}(P)}

On solving, we get  P^{-1}=\frac{1}{2}\left[\begin{array}{ccc} 0 & -2 & 1 \\ -2 & -2 & 0 \\ -2 & -2 & -1 \end{array}\right]

Putting in the Diagonalization of Matrix equation, we get 

\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}        

 

Matlab




% MATLAB Implementation for
% Diagonalization of a Square Matrix:
clear all
clc      
disp("MATLAB Implementation for Diagonalization
 of a Square Matrix | GeeksforGeeks")
A = input("Enter a matrix A : ");
[P , D] = eig(A);
D1 = inv(P)*(A)*(P);
disp("Diagonal form 'D' of Input Matrix 'A' is:")
disp(D1)

 Output:

For the Matrix: A = \begin{bmatrix} 1 &  3\\ 3 &  9\\ \end{bmatrix}

 

 For the Matrix:  A = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 2 & 1 \\ 2 &  2&  3\\ \end{bmatrix}

 


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