Diagonalization - created from Mind Map

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For the First Case

Eigenvector Problem

Matrix A is diagonalizable when A is similar to a diagonal matrix

Matrix B is said similar to matrix A if there is a nonsingular matrix P such that

Prodcedure for Diagonalizing An n x n Matrix A

Symmetrix Matrices & Orthogonal Matrix

A is diagonalizable when there exists an invertible matrix P such that is a diagonal matrix

If A and B are similar n x n matrices, then they have the same eigenvalues

Condition for diagonalization : A is diagonalizable iff it has n linearly independence eigenvectors

If the eigenvalues of an n x n matrix A are all distinct, then A is diagonalizable

Step 1: Find eigenvalue

Step 2: Form matrix P

Step 3: The matrix D = P^(-1)AP will be a diagonal with its principal diagonal entries are the eigenvalues corresponding to eigenvectors.

A square matrix A is symmetric when it is equal to its transpose

A nonsingular matrix A is called orthogonal matrix if A inverse = A transpose

Properties of Symmetric Matrices

1. A is diagonalizable

2. All eigenvalue of A are real

3. If lambda is an eigenvalue of A with multiplicity k, then lambda has k linearly independent eigenvectors. That is eigenspace of lambda has dimension k.

The n x n matrix A is orthogonal iff the column (rows) of A form an orthonormal set of verctors in R^n

Let A be an n x n symmetric matrix . If lambda 1 and lambda 2 are distinct eigenvalues of A, then corresponding eigenvectors x1 and x2 are orthogonal

If A is a symmetric n x n matrix, then there exists an orthogonal matrix P such that is a diagonal matrix

Diagonalization Problem: whether there exist or not an invertible matrix P such that is a diagonal matrix?

If A is an n x n matrix, then A is diagonalizable and A has n linearly independent eigenvectors

Distinct eigenvalues of A -> Distinct eigenvectors of A -> then eigenvectors is a Linearly Independent set

If A is n x n diagonalizable square matrix and is the diagonal matrix, then