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3. A matrix

\mathbf{A} \in \mathbb{C}^{n \times n}

is simple if there exists ...

May 22, 2024

Solution by Steps

step 1

To show that a matrix

\mathbf{A}

is simple if and only if there is a nonsingular matrix

\mathrm{X}

such that

\mathrm{X}^{-1} \mathrm{AX}

is diagonal, we start by assuming

\mathbf{A}

is simple. This means there exists a set of

n

linearly independent eigenvectors for

\mathbf{A}

step 2

Let

\mathbf{X}

be the matrix whose columns are these

n

linearly independent eigenvectors of

\mathbf{A}

. Since the eigenvectors are linearly independent,

\mathbf{X}

is nonsingular

step 3

By definition of eigenvectors, we have

\mathbf{A} \mathbf{x}_i = \lambda_i \mathbf{x}_i

for each eigenvector

\mathbf{x}_i

and corresponding eigenvalue

\lambda_i

step 4

We can write this in matrix form as

\mathbf{A} \mathbf{X} = \mathbf{X} \mathbf{\Lambda}

, where

\mathbf{\Lambda}

is the diagonal matrix with eigenvalues

\lambda_i

on the diagonal

step 5

Multiplying both sides by

\mathbf{X}^{-1}

, we get

\mathbf{X}^{-1} \mathbf{A} \mathbf{X} = \mathbf{\Lambda}

, which shows that

\mathbf{X}^{-1} \mathbf{A} \mathbf{X}

is diagonal

step 6

Conversely, if there exists a nonsingular matrix

\mathbf{X}

such that

\mathbf{X}^{-1} \mathbf{A} \mathbf{X}

is diagonal, let

\mathbf{\Lambda} = \mathbf{X}^{-1} \mathbf{A} \mathbf{X}

step 7

Then

\mathbf{A} = \mathbf{X} \mathbf{\Lambda} \mathbf{X}^{-1}

, and the columns of

\mathbf{X}

are eigenvectors of

\mathbf{A}

corresponding to the eigenvalues on the diagonal of

\mathbf{\Lambda}

step 8

Since

\mathbf{X}

is nonsingular, its columns are linearly independent, showing that

\mathbf{A}

is simple

step 9

To show that the columns of

\mathbf{X}

are right eigenvectors for

\mathbf{A}

and the rows of

\mathbf{X}^{-1}

are left eigenvectors for

\mathbf{A}

, note that

\mathbf{A} \mathbf{X} = \mathbf{X} \mathbf{\Lambda}

implies

\mathbf{A} \mathbf{x}_i = \lambda_i \mathbf{x}_i

step 10

For the left eigenvectors, consider

\mathbf{A}^H \mathbf{y}_i = \lambda_i \mathbf{y}_i

, where

\mathbf{y}_i

are the rows of

\mathbf{X}^{-1}

. Since

\mathbf{X}^{-1} \mathbf{A} \mathbf{X} = \mathbf{\Lambda}

, we have

\mathbf{A}^H = \mathbf{X}^{-H} \mathbf{\Lambda}^H \mathbf{X}^H

step 11

Therefore,

\mathbf{A}^H \mathbf{y}_i = \lambda_i \mathbf{y}_i

shows that the rows of

\mathbf{X}^{-1}

are left eigenvectors of

\mathbf{A}

step 12

For part (b), let

\mathbf{A}

be a simple matrix with eigenvalues

\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}

step 13

There exist right eigenvectors

\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n}

and left eigenvectors

\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}

such that

\mathbf{A} \mathbf{x}_i = \lambda_i \mathbf{x}_i

and

\mathbf{A}^H \mathbf{y}_i = \lambda_i \mathbf{y}_i

step 14

We can express

\mathbf{A}

\mathbf{A} = \mathbf{X} \mathbf{\Lambda} \mathbf{X}^{-1}

, where

\mathbf{X}

is the matrix of right eigenvectors and

\mathbf{\Lambda}

is the diagonal matrix of eigenvalues

step 15

Using the fact that

\mathbf{X}^{-1} = \mathbf{Y}^H

, where

\mathbf{Y}

is the matrix of left eigenvectors, we get

\mathbf{A} = \sum_{i=1}^{n} \lambda_{i} \mathbf{x}_{i} \mathbf{y}_{i}^{H}

Answer

\mathbf{A} = \sum_{i=1}^{n} \lambda_{i} \mathbf{x}_{i} \mathbf{y}_{i}^{H}

Key Concept

Simple Matrix and Eigenvectors

Explanation

A matrix is simple if it has a full set of linearly independent eigenvectors, which allows it to be diagonalized by a nonsingular matrix. The right and left eigenvectors correspond to the columns of the matrix and its inverse, respectively.