Show that (A, B) is controllable if and only if ℭ ℭ^H is invertible.

Name: Problem 7, Chapter 19: Linear Algebra in Action
Uploaded: 2020-07-09T17:36:14-08:00
Duration: 4 min 8 s
Channel: Michael Jacobsen
Description: Show that (A, B) is controllable if and only if ℭ ℭ^H is invertible.

Show that (A, B) is controllable if and only if ℭ ℭ^H is...

Key Concepts

Controllability

Controllability is a fundamental concept in control theory that describes the ability to drive the state of a dynamic system from any initial state to any target state in finite time through the application of an appropriate control input. In the state-space representation, for a system defined by matrices A and B, the system is controllable if the control input can affect every state variable sufficiently to reach every point in the state space.

Controllability Matrix

The controllability matrix is a matrix constructed from the system matrices A and B, typically written as [B, AB, A^2B, …, A^(n?1)B] for an n-dimensional state space. The rank of this matrix is critical; the system is controllable if and only if this matrix has full rank, meaning its columns span the entire state space.

Matrix Invertibility

Invertibility in the context of matrices refers to the ability of a matrix to have an inverse, which is equivalent to the matrix being of full rank. When the product of the controllability matrix and its Hermitian transpose (or conjugate transpose) is invertible, it implies that the original controllability matrix is of full rank, thereby confirming that the system is controllable.

Hermitian Transpose

The Hermitian transpose (or conjugate transpose) of a matrix is obtained by taking the transpose of the matrix and then taking the complex conjugate of each entry. This operation is important in many contexts including the analysis of controllability, as it is used to form symmetric (or Hermitian) matrices such as the Gramian, whose invertibility can be analyzed to draw conclusions about system properties like controllability.

Transcript

00:01 In this exercise, we're assuming that a x equals 0 has only the trivial solution, that's the zero vector, and our goal is to show that a x equals b must have a unique solution for every vector b in rn.

00:15 This is really a two -part proof.

00:17 The first part is going to be an existence proof.

00:23 Here, we're going to show that a x equal b, a x equals b will have a solution, as in it'll be consistent.

00:30 First note that since a times x equals the zero vector has a unique solution, which was the zero vector, we know that a inverse must exist.

00:56 Therefore, if we pick any vector b in rn, if b is in rn, then consider ax equals b.

01:11 This will imply that if we pre -multiply by a inverse in this equation, we'll have a times a inverse x equals a inverse times b, and this then implies that x equals a inverse times b is a solution.

01:31 So this takes care of the existence portion of this proof.

01:35 We know that a x equals b will have a solution, but now we need to show that that solution is unique.

01:43 So this takes us to uniqueness.

01:50 For the uniqueness portion of this proof, we're going to set up the following two solutions.

01:56 Suppose a times u is equal to b and a times v is equal to b for u and v in rn.

02:10 So what we've done is we have assumed we have two solutions to the matrix equation ax equals b.

02:17 Those are solutions.

02:19 U and v.

02:20 Because we're doing a uniqueness proof, our goal is to show of ultimately that u equals v...

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