Diagonalization

The conversion of a matrix into diagonal form is called diagonalization. The eigenvalues of a matrix are clearly represented by diagonal matrices. A Diagonal Matrix is a square matrix in which all of the elements are zero except the principal diagonal elements. Let’s look at the definition, process, and solved examples of diagonalization in this article.

Table of contents:

• Diagonalization of a Matrix
• Diagonalization Theorem
  • Proof
• Diagonalization of 2×2 Matrix
• Diagonalization of 3×3 Matrix
• FAQs

Diagonalization of a Matrix

If there is an invertible n×n matrix C and a diagonal matrix D such that A=CDC^-1, then an n×n matrix A is diagonalizable.

For example,

Hence, we can say, any diagonal matrix D is diagonalizable, as it is similar to itself.

Diagonalization Theorem

If and only if A has n linearly independent eigenvectors, then the n×n matrix A is diagonalizable.

A=CDC^-1 for this example.

Here,

v₁, v₂, …, v_n are the linearly independent Eigenvectors,

λ₁, λ₂, …λ_n are the corresponding Eigenvalues.

Diagonalization Proof

Assume that matrix A has n linearly independent Eigenvectors such as v₁, v₂, …, v_n, having Eigenvalues λ₁, λ₂, …λ_n. Defining “C” as considered above, we can conclude C is invertible using the invertible matrix theorem.

Assume that D = C^-1AC, Hence, A = CDC^-1.

Now, consider multiplying the standard coordinate vectors picks by the elements columns of C, we can have

Ce_i = v_i, and hence e_i = C^-1v_i.

To obtain the columns of D, we multiply by the standard coordinate vectors.

Thus we can say,

De_i = C^-1ACe_i = C^-1Av_i = C^-1λ_iv_i = λ_iC^-1v_i = λ_ie_i

D’s columns are hence multiples of the standard coordinate vectors:

Assume A = CDC^-1, where C has columns v₁, v₂,…, v_n, and D is diagonal with diagonal entries λ₁, λ₂,…, λ_n. C’s columns are linearly independent since it is invertible. We need to demonstrate that v_i is an eigenvector of A with eigenvalue λ_i. Because the standard coordinate vector e_i is an eigenvector of D with eigenvalue λ_i we can write:

Av_i = CDC^-1V_i = CDe_i = Cλ_ie_i = λ_iCe_i = λ_iv_i.

Hence, we can conclude that if

If an n×n matrix A has n different eigenvalues λ₁, λ₂,…, λ_n, then a selection of matching eigenvectors v₁, v₂,…, v_n is inherently linearly independent.

In other words, an n×n matrix with unique Eigenvalues is diagonalizable.

Diagonalization of 2×2 Matrix

Go through the below example to understand the procedure for diagonalizing the 2×2 matrix.

Example 1:

Diagonalize the given 2×2 matrix:

Solution:

First, calculate the characteristic polynomial to find the Eigenvalues and Eigenvectors.

Hence, f(λ)= λ² – Tr(A)λ + det(A) = λ² – λ – 2 = (λ+1)(λ-2)

Thus, the Eigenvalues are -1 and 2.

Now, we have to calculate the Eigenvector for every Eigenvalue.

Thus, if λ₁= -1,

Now, the parametric form is x = -y, and hence

Similarly, the Eigenvector for the Eigenvalue λ₂= 2 is given by:

Hence, the parametric form is x = y, and hence

Therefore, the Eigenvectors v₁ and v₂ are linearly independent and according to the diagonalization theorem, we can write

A = CDC^-1 for

In case, if we choose λ=2, then we can get

A = C’D(C’)^-1 for

Diagonalization of 3×3 Matrix

For a better understanding of how to diagonalize a 3×3 matrix, look at the example below.

Example 2:

Diagonalize the given matrix:

Solution:

Firstly, find the characteristic polynomial by expanding the cofactors of 3rd column to calculate the Eigenvalue First, calculate the characteristic polynomial by expanding the cofactors along the 3rd column, to find Eigenvalues and Eigenvectors.

f(λ) = (1-λ)(λ²-3λ+2) =-(λ-1)²(λ-2).

Hence, the Eigenvalues are 1 and 2.

Thus, the Eigenvector for the Eigenvalue λ₁ = 1.

Thus, the parametric vector form is

Therefore, the basis for the 1-Eigenspace is

Where,

Similarly, compute the Eigenspace for λ₂ = 2.

Therefore, the parametric form is x = 3z, and y =2z, then the Eigenvector for the Eigenvalue 2 is

Hence, the Eigenvectors v₁, v₂ and v₃ are linearly independent.

Here, v₁ and v₂ form the basis of 1-Eigenspace, whereas v₃ does not belong to 1-Eigenspace, as its Eigenvalue is 2.

Hence, from the diagonalization theorem, we can write

A = CDC^-1, for