Problem 11 - Differential Equations and Linear Algebra

Question

Let \(A(t)\) be an \(n \times n\) matrix function. Prove that the set of all solutions \(\mathbf{x}\) to the system \(\mathbf{x}^{\prime}(t)=A(t) \mathbf{x}(\mathrm{t})\) is a subspace of \(V_{n}(I)\)

Step-by-Step Solution

Answer

To show that the set of all solutions \(\mathbf{x}\) to the system \(\mathbf{x}'(t)=A(t)\mathbf{x}(t)\) is a subspace of \(V_n(I)\), we demonstrated that it is closed under vector addition, scalar multiplication, and contains the zero vector. By showing that the sum of two solutions, the scalar multiple of a solution, and the zero vector are all solutions to the system, we can conclude that the set of all solutions is a subspace of \(V_n(I)\).

Step 1: Rewrite the System

Rewrite the given system x'(t) = A(t)x(t) in terms of a matrix differential equation.

Step 2: Consider Two Solutions x1(t) and x2(t)

Let x1(t) and x2(t) be two solutions of the system x'(t) = A(t)x(t), i.e., x1'(t) = A(t)x1(t) and x2'(t) = A(t)x2(t).

Step 3: Check if Set is Closed under Vector Addition

Consider the sum of these solutions, y(t) = x1(t) + x2(t). Differentiate y(t) with respect to t to obtain y'(t): y'(t) = x1'(t) + x2'(t) = A(t)x1(t) + A(t)x2(t) = A(t)(x1(t) + x2(t)) = A(t)y(t). Since it is shown that the derivative y'(t) is equal to A(t)y(t), the sum of the two solutions, y(t), is also a solution. Therefore, the set of solutions is closed under vector addition.

Step 4: Check if Set is Closed under Scalar Multiplication

Consider a scalar, c, multiplied by the solution x1(t), z(t) = cx1(t). Differentiate z(t) with respect to t to obtain z'(t): z'(t) = c(x1'(t)) = c(A(t)x1(t)) = A(t)(cx1(t)) = A(t)z(t). Since it is shown that the derivative z'(t) is equal to A(t)z(t), the scalar multiple of the solution, z(t), is also a solution. Therefore, the set of solutions is closed under scalar multiplication.

Step 5: Check if Set Contains the Zero Vector

Consider the zero vector in V_n(I), denoted as 0. Since A(t) is an n x n matrix, the product A(t)0 is equal to 0. The derivative of the zero vector with respect to t is also 0: 0'(t) = A(t)0. Thus, the zero vector is a solution of the system x'(t) = A(t)x(t), and the set of solutions contains the zero vector.

Step 6: Conclude that the Set of Solutions is a Subspace of Vn(I)

Since we've demonstrated that the set of solutions to the system x'(t) = A(t)x(t) is closed under vector addition, scalar multiplication, and contains the zero vector, we conclude that the set of solutions is a subspace of V_n(I).

Key Concepts

Matrix Differential Equation

Understanding matrix differential equations is fundamental when exploring systems of linear equations that change over time. Similar to ordinary differential equations, which relate functions to their derivatives, matrix differential equations do the same for vector-valued functions.

For example, let \(A(t)\) be a time-dependent \(n \times n\) matrix, and \(\mathbf{x}(t)\) be a vector-valued function. The equation \(\mathbf{x}'(t)=A(t) \mathbf{x}(t)\) is a matrix differential equation, where \(\mathbf{x}'(t)\) denotes the derivative of \(\mathbf{x}(t)\) with respect to time \(t\).

In the context of our problem, solutions to this equation are vector functions that fit into the system \(\mathbf{x}'(t)=A(t)\mathbf{x}(t)\). Each of these solutions can be thought of as a trajectory in an \(n\)-dimensional space, defined by how the vector \(\mathbf{x}(t)\) evolves over time.

Vector Addition Closure

A key property of any subspace is closure under vector addition. This means that for any two vectors \(\mathbf{x}_1\) and \(\mathbf{x}_2\) in the set, their sum \(\mathbf{y} = \mathbf{x}_1 + \mathbf{x}_2\) must also be in the set.

In our exercise, we considered two solutions, \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\), of the matrix differential equation. By showing that the sum \(\mathbf{y}(t)\) of these two solutions is itself a solution to the same differential equation, we demonstrate that the solution set is closed under vector addition.

\textbf{Closure Demonstration:} Differentiating the sum gives us \(\mathbf{y}'(t) = \mathbf{x}_1'(t) + \mathbf{x}_2'(t)\), and using the fact that both \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are solutions, we get that \(\mathbf{y}'(t) = A(t)\mathbf{y}(t)\). Thus, the sum complies with the original matrix differential equation, ensuring that the property of vector addition closure is satisfied.

Scalar Multiplication Closure

Along with vector addition, scalar multiplication closure is necessary for a set to be considered a subspace. A set is closed under scalar multiplication if, for any vector \(\mathbf{x}\) in the set and any scalar \(c\), the product \(\mathbf{z} = c\mathbf{x}\) also belongs to the set.

In our step-by-step solution, we showed that if \(\mathbf{x}_1(t)\) is a solution to the matrix differential equation, then scaling it by any scalar \(c\) results in another solution \(\mathbf{z}(t) = c\mathbf{x}_1(t)\). This conclusion is reached by differentiating \(\mathbf{z}(t)\) to find \(\mathbf{z}'(t)\) and observing that it still satisfies the equation \(\mathbf{z}'(t) = A(t)\mathbf{z}(t)\) - proving the scalar multiplication closure.

Zero Vector

The presence of the zero vector in a set is a third essential criterion for it to be a subspace. The zero vector is the additive identity in vector spaces, meaning that adding it to any vector in the space doesn't change the original vector.

Our analysis addressed this criterion by taking the zero vector \(\mathbf{0}\) in an \(n\)-dimensional space and showing that it is a solution to the matrix differential equation \(\mathbf{x}'(t) = A(t)\mathbf{x}(t)\). Since \(A(t)\mathbf{0}\) is always \(\mathbf{0}\) and the derivative of the constant vector \(\mathbf{0}\) over time is also \(\mathbf{0}\), the zero vector is indeed a solution to the equation. Consequently, the solution set includes this special vector, fulfilling the zero vector criterion for a subspace.