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Q7.6-26 E

Expert-verified
Found in: Page 267

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Consider an invertible n × n matrix A such that the zero state is a stable equilibrium of the dynamical system $\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{+}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{A}}^{\mathbf{T}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}$ What can you say about the stability of the systems.$\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{+}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{A}}^{\mathbf{T}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}$

The given value is stable

See the step by step solution

## Define eigen value

Eigenvalues can be used to determine whether a fixed point (also known as an equilibrium point) is stable or unstable.

## Finding the stability of the given equation

If the system that involves ${\mathrm{A}}^{\mathrm{T}}$ is stable, then all of its eigenvalues are inside the unit circle. since and ${\mathrm{A}}^{\mathrm{T}}$ have the exact same eigenvalues then this also applies for ${\mathrm{A}}^{\mathrm{T}}$ meaning that the system that involves ${\mathrm{A}}^{\mathrm{T}}$ is stable.