Matrix Diagonalization

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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 λ_{1, 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, λ_{i ->} eigen value. X_{i ->} corresponding eigen vector.

Step 4: Create the modal matrix P.

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

Here, all the eigenvectors till X_i 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^-1and then putting values in diagonalization of a matrix equation. [D = P^-1AP]

We do Step 6 to find out which eigenvalue will replace λ₁, λ_2,and λ₃in 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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Last Updated : 27 Apr, 2022