Question

Suppose the columns of an \(n \times n\) matrix \(Y\) are solutions of the \(n \times n\) system \(\mathbf{y}^{\prime}=A \mathbf{y}\) and \(C\) is an \(n \times n\) constant matrix. (a) Show that the matrix \(Z=Y C\) satisfies the differential equation \(Z^{\prime}=A Z\). (b) Show that \(Z\) is a fundamental matrix for \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\) if and only if \(C\) is invertible and \(Y\) is a fundamental matrix for \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\).

Short answer

To summarize, we have proven two statements about the matrix Z: (a) We have shown that the matrix Z = YC satisfies the differential equation Z' = AZ. (b) We have shown that Z is a fundamental matrix for the system y'=A(t)𝐲 if and only if C is invertible and Y is a fundamental matrix for y'= A(t) 𝐲. Our proof consisted of first differentiating the matrix Z and showing that it's equal to AZ and then proving both directions of the second claim by analyzing the properties of invertible, fundamental matrices, and their determinants.

Step-by-step solution

(a) Proving that \(Z' = AZ\)

Given that the columns of the matrix \(Y\) are the solutions of the system \(\mathbf{y}' = A\mathbf{y}\). So, we have \(Y' = AY\). We need to show that the matrix \(Z = YC\) satisfies the differential equation \(Z' = AZ\). Let's find the differential of \(Z\). First, differentiate the matrix \(Z\) with respect to \(t\): $$ Z'= \frac{d}{dt}(YC) = Y'\cdot C $$ Since \(Y' = AY\), we can substitute this expression: $$ Z' = AY\cdot C $$ Recall that matrix multiplication is associative, so we can rewrite this expression as: $$ Z' = A(YC) $$ And since \(Z=YC\), we have: $$ Z' = AZ $$ Thus, the matrix \(Z\) satisfies the differential equation \(Z' = AZ\).

(b) Proving if and only if statements about fundamental matrices

We have to show that \(Z\) is a fundamental matrix for \(\mathbf{y}'=A(t)\mathbf{y}\) if and only if \(C\) is invertible and \(Y\) is a fundamental matrix for \(\mathbf{y}'= A(t) \mathbf{y}\). 1. Assume that \(C\) is invertible and \(Y\) is a fundamental matrix. We want to show that \(Z = YC\) is also a fundamental matrix. A fundamental matrix has to satisfy two conditions: its determinant is nonzero, and its columns form a linearly independent set of solutions to the given system of equations. First, let's show that \(Z\) has a nonzero determinant: $$ \det(Z) = \det(YC) = \det(Y) \cdot \det(C) $$ As \(C\) is invertible, its determinant must be nonzero: \(\det(C)\neq 0\). Similarly, as \(Y\) is a fundamental matrix, its determinant must be nonzero as well: \(\det(Y)\neq 0\). Therefore, \(\det(Z) \neq 0\). Now, let's show that the columns of \(Z\) form a linearly independent set of solutions. Since \(Y\) is a fundamental matrix, its columns form a linearly independent set of solutions. Definition of \(Z\) gives that the columns of \(Z\) are linear combinations of the columns of \(Y\) using the columns of the invertible matrix \(C\). As the columns of \(Y\) are linearly independent and \(C\) is an invertible matrix, it follows that the columns of \(Z\) are linearly independent as well. Thus, \(Z\) is a fundamental matrix for \(\mathbf{y}'=A(t)\mathbf{y}\). 2. Conversely, assume that \(Z\) is a fundamental matrix for \(\mathbf{y}'=A(t)\mathbf{y}\). We want to show that \(C\) must be invertible and \(Y\) must be a fundamental matrix. The determinant of \(Z\) is nonzero since it's a fundamental matrix, and since \(Z = YC\), we have: $$ \det(Z) = \det(YC) = \det(Y) \cdot \det(C) $$ Since \(\det(Z) \neq 0\), it must be the case that both \(\det(Y)\) and \(\det(C)\) are nonzero. Therefore, \(C\) is invertible and \(Y\) has a nonzero determinant. As \(Z=AY\) and \(Z\) is a fundamental matrix, the columns of \(Z\) form a linearly independent set of solutions. Since the columns of \(Z\) are linear combinations of the columns of \(Y\) using the invertible matrix \(C\), the columns of \(Y\) must be linearly independent as well. Therefore, \(Y\) is a fundamental matrix for \(\mathbf{y}'= A(t) \mathbf{y}\). In conclusion, we have shown that \(Z\) is a fundamental matrix for \(\mathbf{y}'=A(t)\mathbf{y}\) if and only if \(C\) is invertible and \(Y\) is a fundamental matrix for \(\mathbf{y}'= A(t) \mathbf{y}\), completing the exercise.

Key concepts

Linear Differential Equations

Linear differential equations form the foundation for understanding a range of physical phenomena and mathematical concepts. In essence, a linear differential equation involves an unknown function and its derivatives. It is called 'linear' because the unknown function and its derivatives appear to the first power and are not multiplied together.

One of the simplest forms is the first-order linear differential equation, expressed as \( \frac{dy}{dt} + p(t)y = q(t) \), where \( p(t) \) and \( q(t) \) are functions of \( t \) and \( y \) is the function of \( t \) we aim to find. The beauty of linear differential equations lies in their predictability and the existence of well-defined methods for their solutions, such as integrating factors.

In the context of systems of linear differential equations, we extend these principles to multiple functions and their derivatives, resulting in a set of simultaneous equations. Understanding these systems is crucial for fields like engineering, physics, and economics where multiple interacting variables are the norm.

Matrix Differential Equations

When we elevate from single linear differential equations to systems, matrix differential equations become exceedingly useful. These are equations where the solutions are vector functions rather than scalar functions. A standard form for a matrix differential equation is \( \mathbf{y}' = A\mathbf{y} \), where \( \mathbf{y} \) is a vector of functions and \( A \) is a matrix of coefficients which can be constants or functions of independent variable \( t \).

In the matrix form, a single equation embodies the entire system, packing a powerful punch. Each entry in matrix \( A \) dictates how each function in vector \( \mathbf{y} \) interacts with one another, making it easier to handle a system with multiple interrelated components.

The concept of a fundamental matrix arises in this context as well, which is an \( n \times n \) matrix solution to the matrix differential equation. Each column of a fundamental matrix is an independent solution to the system. The existence of a fundamental matrix greatly simplifies solving matrix differential equations, as it can be used to construct general solutions.

Invertible Matrix

The concept of an invertible matrix, also known as a non-singular or nonsingular matrix, plays a pivotal role in understanding matrix equations and linear algebra. An \( n \times n \) matrix is invertible if there exists another \( n \times n \) matrix \( C \) such that \( AC = CA = I \), where \( I \) is the identity matrix. In other words, the matrix \( C \) is the inverse of \( A \) and is denoted by \( A^{-1} \).

The presence of an invertible matrix ensures that we can perform operations akin to division by a number, except in the realm of matrices. This becomes particularly useful in solving systems of linear equations, where an invertible coefficient matrix allows for finding unique solutions. For a matrix to be invertible, it must have a nonzero determinant and full rank, meaning all rows and columns are linearly independent.

Nonetheless, when dealing with differential equations, the invertibility of a matrix has deep implications. It allows transformations that can change the form of the solution but not its essence, much like altering the frame of reference in physics without changing the physical phenomenon's characteristics. The invertibility of a transformation matrix \( C \) in our exercise is vital to prove the existence of a fundamental matrix, emphasizing the interconnectedness of linear algebra with differential equations.