Question

Suppose \(Y\) and \(Z\) are fundamental matrices for the \(n \times n\) system \(\mathbf{y}^{\prime}=A(t) \mathbf{y} .\) Then some of the four matrices \(Y Z^{-1}, Y^{-1} Z, Z^{-1} Y, Z Y^{-1}\) are necessarily constant. Identify them and prove that they are constant.

Short answer

a) \(YZ^{-1}\) b) \(Y^{-1}Z\) c) \(Z^{-1}Y\) d) \(ZY^{-1}\)

Step-by-step solution

Finding the derivatives of the combinations of Y and Z

Let's derive \(YZ^{-1}\), \(Y^{-1}Z\), \(Z^{-1}Y\), and \(ZY^{-1}\) with respect to t. 1) \((YZ^{-1})'= Y'Z^{-1} - YZ^{-1}(Z^{-1})'Z^{-1}\) 2) \((Y^{-1}Z)' = -Y^{-1}Y'Y^{-1} Z + Y^{-1}Z'\) 3) \((Z^{-1}Y)' = -Z^{-1}Z'Z^{-1} Y + Z^{-1}Y'\) 4) \((ZY^{-1})'= Z'Y^{-1} - ZY^{-1}(Y^{-1})'Y^{-1}\) Now, we know that the given system can be written as \(Y'=A(t)Y\) and \(Z'=A(t)Z\). Let's use this to simplify the expressions above.

Substitute Y' and Z' with A(t)Y and A(t)Z respectively

Substitute these expressions into the derivatives calculated above: 1) \((YZ^{-1})'= A(t)Y Z^{-1} - YZ^{-1}(Z^{-1})'(Z^{-1} Z)\) 2) \((Y^{-1}Z)' = -Y^{-1}A(t)Y Y^{-1} Z + Y^{-1}A(t)Z\) 3) \((Z^{-1}Y)' = -Z^{-1}A(t)Z Z^{-1} Y + Z^{-1}A(t)Y\) 4) \((ZY^{-1})'= A(t)Z Y^{-1} - ZY^{-1}(Y^{-1})'(Y^{-1}Y)\)

Identify constant matrices

Now, we need to find which of these expressions have the derivative equal to zero. 1) \((YZ^{-1})' = A(t)YZ^{-1} - A(t) = 0 \implies A(t)(YZ^{-1}-I) = 0\) 2) \((Y^{-1}Z)' = -Y^{-1}A(t)Y Y^{-1} Z + Y^{-1}A(t)Z = Y^{-1}A(t)(I - YZ^{-1})\) 3) \((Z^{-1}Y)' = -Z^{-1}A(t)Z Z^{-1} Y + Z^{-1}A(t)Y = Z^{-1}A(t)(Y-Z^{-1})\) 4) \((ZY^{-1})' = A(t)ZY^{-1} - A(t) = 0 \implies A(t)(ZY^{-1}-I) = 0\) From the expressions above, we can see that \((YZ^{-1})'\) and \((ZY^{-1})'\) are zero. Therefore, the matrices \(YZ^{-1}\) and \(ZY^{-1}\) are constant matrices.

Conclusion

The matrices \(YZ^{-1}\) and \(ZY^{-1}\) are necessarily constant matrices in the given system.

Key concepts

System of Linear Differential Equations

When working with a system of differential equations, we often encounter a set of multiple equations that describe various interconnected rates of change. In linear algebra, a system of linear differential equations typically involves vectors of functions and their derivatives. These systems can be conveniently represented using matrix notation, which significantly simplifies the solution process.

For instance, when confronted with the equation \(\mathbf{y}^{\'}=A(t)\mathbf{y}\), we are dealing with a first-order linear system, where \(\mathbf{y}\) is a vector of functions, \(A(t)\) is a matrix of coefficients that may vary with time, and the prime denotes differentiation with respect to time. Fundamental matrices, such as \(Y\) and \(Z\) mentioned in our exercise, are special solutions that span the space of all solutions to our differential equation system. Due to their importance in constructing general solutions, grasping the concept of fundamental matrices is key for understanding the behavior of linear systems of differential equations.

Matrix Differentiation

Matrix differentiation, while less frequently discussed than differentiation in the context of single-variable calculus, is an important tool when analyzing systems of linear differential equations. Differentiation of matrices operates under specific rules that extend our understanding from scalar functions to matrices.

As seen in the exercise above, the calculations involved in differentiating products of matrices, such as \(YZ^{-1}\), use a matrix analogue of the product rule from single-variable calculus. It's essential to remember that since matrices do not generally commute—that is, \(AB\) may not equal \(BA\)—the order of multiplication matters significantly. Understanding and applying these differentiation rules is crucial to simplifying expressions and solving matrix equations within the system of differential equations.

Constant Coefficient Matrices

In the context of differential equations, 'constant coefficient matrices' often refers to matrices in which the coefficients of the system do not change over time. This constancy provides a considerable simplification when solving differential equations, as it means that the coefficients can be treated as constants when integrating or differentiating.

This property was explored in the problem under study, where the solution process involved identifying matrices that remain constant over time. When the rates of change of certain matrix products, such as \(YZ^{-1}\) and \(ZY^{-1}\), yield zero, those matrix products are proven to be constants within the system. A good understanding of constant coefficient matrices is not only vital for analyzing the stability and behavior of solutions but also plays a major role in applications such as control theory and signal processing.