Let A and B be square matrices of order n. The matrix B is said to be similar to A if there exists an invertible matrix P such that
B = P-1 A P
This relation is called a matrix similarity transformation. Similar matrices represent the same linear transformation with respect to different choices of basis.
Diagonalisation of a matrix is the process of converting a square matrix A into a diagonal matrix D by a similarity transformation. That is, if there exists an invertible matrix P such that
D = P-1 A P
then A is said to be diagonalizable, and the matrix P is often called a modal matrix.
Modal matrix: It is an n × n matrix whose columns are eigenvectors of A. The modal matrix is used to diagonalise A when a full set of linearly independent eigenvectors exists.
If the columns of P are eigenvectors of A, and the corresponding eigenvalues are placed on the diagonal of D in the same order, then the similarity relation A = P D P-1 (equivalently D = P-1 A P) holds. Diagonal matrices are simpler to work with; many matrix computations (powers, exponentials, solving systems) become easier when matrices are diagonal.



If A = P D P-1 with D diagonal, then columns of P are eigenvectors of A and D contains the corresponding eigenvalues on its diagonal.
Conversely, if A has n linearly independent eigenvectors v1,...,vn, form P = [v1 ... vn] and let D be the diagonal matrix with diagonal entries λ1,...,λn where A vi = λi vi. Then AP = P D and hence P-1 A P = D.
Problem Statement: Assume a 3×3 square matrix A having the following values:

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

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


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



On solving, we get the following equation
x3 = 0 (x1)x1 + x2 = 0


Similarly, for λ = 2

and for

similarly

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

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.



Since det(P) ≠ 0 ⇒ Matrix P is invertible
we know that

On solving, we get

Putting in the Diagonalization of Matrix equation, we get

Compute the characteristic polynomial:
Form the matrix A - λI and compute det(A - λI) = 0 to obtain the eigenvalues.
For each eigenvalue λi, solve the homogeneous linear system (A - λiI) X = 0 to obtain the eigenspace and basis eigenvectors corresponding to λi.
Collect one independent eigenvector from each eigenspace until you have three independent vectors (for a 3×3 matrix). Place these eigenvectors as columns of P in the same order as their eigenvalues will be placed on the diagonal of D.
Verify that det(P) ≠ 0 to ensure P is invertible.
Compute P-1 (for a 3×3 matrix use the adjoint/determinant method or Gaussian elimination) and form D = P-1 A P.
The diagonal entries of D are the eigenvalues placed in the order corresponding to the columns of P. The final D is thus a diagonal matrix with those eigenvalues on its diagonal.
...
) in the example shows the characteristic polynomial calculation, eigenvector computations, the modal matrix P, the inverse P-1, and the resulting diagonal matrix D. These illustrate the standard computational route described in the procedure.Summary - Diagonalisation reduces a matrix to a diagonal form by a similarity transformation using a modal matrix whose columns are eigenvectors. A matrix is diagonalizable iff it has a full set of linearly independent eigenvectors; real symmetric matrices are always diagonalizable by orthogonal matrices. Diagonalisation is a central tool in many engineering computations including modal analysis, solution of linear differential systems, and efficient computation of matrix powers.
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