Matrix Diagonalization

Matrix diagonalization is the process of reducing a square matrix into its diagonal form using a similarity transformation. This process is useful because diagonal matrices are easier to work with, especially when raising them to integer powers.

Not all matrices are diagonalizable. A matrix is diagonalizable if it has no defective eigenvalues, meaning each eigenvalue's geometric multiplicity is equal to its algebraic multiplicity.

Matrix Similarity Transformation

Let A and B be two matrices of order n. Matrix B is considered similar to A if there exists an invertible matrix P such that:

B = P^-1 A P

This transformation is known as Matrix Similarity Transformation. Similar matrices have the same rank, trace, determinant, and eigenvalues. Additionally, the eigenvalues of similar matrices maintain their algebraic and geometric multiplicities.

Diagonalization of a Matrix

Diagonalization of a matrix refers to the process of transforming any matrix A into its diagonal form D. According to the similarity transformation, if A is diagonalizable, then:

D=P^{-1} A P 
where D is a diagonal matrix and P is a modal matrix.

A modal matrix is an n × n matrix consisting of the eigenvectors of A. It is essential 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 to Diagonalize a Matrix

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 3x3 square matrix.
I -> identity matrix of size 3x3.
λ -> eigen value.

Step 3: Compute the corresponding eigen vectors using the equation given below.
At λ=i
[A - \lambda I] X_i = 0
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 3x3 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 \end{array} 
⟹\lambda=1,2,3

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

At λ=1
(A - (1) I) X_{1}=0

\Longrightarrow \begin{bmatrix} 1-1 & 0 & -1 \\ 1 & 2-1 & 1 \\ 2 & 2 & 3-1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

\Longrightarrow \begin{bmatrix} 0 & 0 & -1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

On solving, we get the following equations:

x_{3}=0(x_{1})
x_{1}+x_{2}=0 \Longrightarrow x_{2}=-x_{1}

\therefore X_{1}= \begin{bmatrix} x_{1} \\ -x_{1} \\ 0(x_{1}) \end{bmatrix}

\Longrightarrow X_{1}= \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}

Similarly, for λ=2:

X_{2}= \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}

And for λ=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^-1 AP]

We do Step 6 to find out which eigenvalue will replace λ₁, λ_2, 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)=[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 Implementation for Diagonalization of a Square Matrix

% 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}

The Diagonalization is Not Unique

If a matrix A is diagonalizable, there is no unique way to diagonalize it.

1. The order of eigenvalues in the diagonal matrix D can be changed.
2. A column in P can be replaced with a scalar multiple of itself, as it remains an eigenvector for the same eigenvalue.
3. If an eigenvalue is repeated, different bases for its eigenspace can be chosen.

Example

in the previous example, we could have defined:

D = \begin{bmatrix} \lambda_2 & 0 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}

and
P = \begin{bmatrix} x_2 & x_1 & x_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 2 & 1 \\ 2 & 2 & 3 \end{bmatrix}

Another possibility would have been to choose:

x_2 = \begin{bmatrix} -1 \\ -1 \\ -2 \end{bmatrix}

and
P = \begin{bmatrix} x_2 & x_1 & x_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & -1 \\ -1 & 2 & 1 \\ -2 & 2 & 3 \end{bmatrix}

This shows that diagonalization is not unique since we can change eigenvectors by scalar multiples or reorder them while maintaining the same eigenvalues.

Inverse Matrix

Once a matrix has been diagonalized, computing its inverse (if it exists) becomes straightforward.

In fact, we have:
A^{-1} = (PDP^{-1})^{-1} =P D^{-1} P^{-1}

where:
D^{-1} = \text{diag} \left(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \dots, \frac{1}{\lambda_n} \right)

provided none of the eigenvalues λ are zero.

Practice Problems on Matrix Diagonalization

Problem 1: Check if a Matrix is Diagonalizable
Given the matrix: A = \begin{bmatrix} 4 & 1 \\ 6 & -1 \end{bmatrix}
Find the eigenvalues and eigenvectors. Determine if A is diagonalizable.

Problem 2: Compute the Diagonal Form
For the matrix: A = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix}
Find the diagonal matrix D and the matrix P such that A= P D P⁻¹.

Problem 3: Find the Inverse Using Diagonalization
Given the matrix: A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}
Diagonalize A, and use it to compute A⁻¹, if it exists.

Problem 4: Repeated Eigenvalues Case
Consider the matrix: A = \begin{bmatrix} 6 & -2 \\ 2 & 2 \end{bmatrix}
Check if it is diagonalizable, given that it has a repeated eigenvalue.

Problem 5: Diagonalizing a 3×3 Matrix
For the matrix: A = \begin{bmatrix} 7 & -4 & 4 \\ -4 & 8 & -4 \\ 4 & -4 & 7 \end{bmatrix}
Find the eigenvalues and eigenvectors. Determine P and D for diagonalization.