Question

Show that \(Y\) is a fundamental matrix for the system \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\) if and only if \(Y^{-1}\) is a fundamental matrix for \(\mathbf{y}^{\prime}=-A^{T}(t) \mathbf{y},\) where \(A^{T}\) denotes the transpose of \(A\). HINT: See Exercise 11

Short answer

Question: Show that a matrix Y is a fundamental matrix for the given system of differential equations if and only if its inverse Y^{-1} is a fundamental matrix for the system with the transpose of the original matrix. Answer: To prove that Y is a fundamental matrix for the given system if and only if Y^{-1} is a fundamental matrix for the transpose system, we need to show the relationship between the Wronskians of Y and the transpose of Y^{-1}. Using the relationship between the transpose and inverse of a matrix and finding the Wronskians, we were able to show that W[Z^T] = det(Y^{-1}) * W[Y]. Since the determinant of Y^{-1} cannot be zero, we then demonstrated that the Wronskians of the two systems are connected, proving the required statement: W[Z^T] ≠ 0 if and only if W[Y] ≠ 0.

Step-by-step solution

Background Relationship between Transpose and Inverse

Recall the relationship between the transpose of a matrix \(A\) and the inverse of its transpose, given by \((A^{-1})^{T} = (A^{T})^{-1}\).

Fundamental Matrix Definitions

By definition, a matrix \(Y\) is a fundamental matrix for the given system if the Wronskian \(W[Y]\) has a non-zero determinant, i.e., \(\det(Y)\neq 0\). Similarly, \(Y^{-1}\) is a fundamental matrix if the Wronskian \(W[Y^{-1}]\) has a non-zero determinant = \(\det(Y^{-1})\neq 0\).

Compute the Wronskians of Y and Y^{-1}

Compute the Wronskians of \(Y\) and \(Y^{-1}.\) Remember that the Wronskian of a set of functions is the determinant of the matrix formed by those functions and their derivatives. The Wronskian test states that if the Wronskian of a set of functions is non-zero, then the functions are linearly independent, and hence form a fundamental set of solutions. Let \(Y = \begin{bmatrix}y_{1}(t) & y_{2}(t) & ... & y_{n}(t)\end{bmatrix}\). Then the Wronskian is: $$W[Y] = \det\begin{bmatrix}y_{1}(t) & y_{1}^{\prime}(t) & ... & y_{n}^{\prime}(t)\\ y_{2}(t) & y_{2}^{\prime}(t) & ... & y_{n}^{\prime}(t)\\ ... & ... &... &...\\ y_{n}(t) & y_{n}^{\prime}(t) & ... & y_{n}^{\prime}(t) \end{bmatrix}$$ Now, let \(Z = Y^{-1} =\begin{bmatrix}z_{1}(t) & z_{2}(t) & ... & z_{n}(t)\end{bmatrix}\). Then the Wronskian of its transpose is: $$W[Z^{T}] = \det\begin{bmatrix}z_{1}(t) & z_{1}^{\prime}(t) & ... & z_{n}^{\prime}(t)\\ z_{2}(t) & z_{2}^{\prime}(t) & ... & z_{n}^{\prime}(t)\\ ... & ... &... &...\\ z_{n}(t) & z_{n}^{\prime}(t) & ... & z_{n}^{\prime}(t) \end{bmatrix}$$

Show the Relationship between W[Y] and W[Z^T]

To show that if and only if \(Y\) is a fundamental matrix for the original system, \(Y^{-1}\) is a fundamental matrix for the transpose system, we need to demonstrate the relationship between \(W[Y]\) and \(W[Z^T],\) and show that one cannot exist without the other. We can do that using the relationship \((A^{-1})^T = (A^T)^{-1}\). Substituting \(Z = Y^{-1}\) gives us \((Y^{-1})^T=Y^{-T}\). Now from the above formula and the equation \(\mathbf{y}^{\prime}=A(t)\mathbf{y}\) we get that: $$W[Z^T] = \det(Y^{-1} Y^{\prime}-Y^{-1}(A(t)Y)) = \det(Y^{-1}(Y^{\prime}-A(t)Y)) = \det(Y^{-1}) \cdot \det(Y^{\prime}-A(t)Y)$$ Now note that \(W[Y] = \det(Y^{\prime}-A(t)Y)\), so we can write: $$W[Z^T] = \det(Y^{-1}) \cdot W[Y]$$ Since \(\det(Y^{-1})\neq 0\) (by definition as a fundamental matrix), we have shown the relationship between the Wronskians, hence proving the required statement: $$W[Z^T] \neq 0 \iff W[Y] \neq 0$$

Key concepts

Fundamental Matrix

In differential equations, a fundamental matrix is a crucial concept when solving a system of linear differential equations. A fundamental matrix is made up of solutions, where each column represents an individual solution to the differential system. For a system given by \(\mathbf{y}' = A(t) \mathbf{y}\), the fundamental matrix often denoted by \(Y(t)\) has the property that its determinant, known as the Wronskian \(W[Y]\), is not zero at any point in time.

This implies linear independence among the columns of the matrix \(Y(t)\). Essentially, the fundamental matrix provides a complete set of solutions to the system, allowing one to express the general solution of the system as combinations of these column solutions.

Therefore, verifying whether \(Y\) satisfies the condition \(\det(Y) eq 0\) is a key step in confirming its status as a fundamental matrix.

Wronskian

The Wronskian is a mathematical determinant useful in understanding the linear independence of a set of solutions to a system of differential equations. For a fundamental matrix \(Y(t)\), the Wronskian \(W[Y]\) is computed as the determinant of the matrix formed by the functions \(y_i(t)\) and their derivatives.

The importance of the Wronskian lies primarily in its ability to indicate linear independence through its value:

• If \(W[Y]eq 0\), the solutions are linearly independent. This means none of the solutions can be written as a combination of the others.
• If \(W[Y] = 0\), it suggests potential linear dependence, implying one solution can be derived from others.

The role of the Wronskian as a determinant in determining whether a set of function solutions is a valid basis for all possible solutions of the system is fundamental in the embodiment of the fundamental matrix.

Linear Independence

Linear independence is a critical property for solutions in differential equations, ensuring that no solution can be expressed as a linear combination of others in the set. In the context of a fundamental matrix, linear independence guarantees that the matrix provides a complete, unique set of solutions.

This concept is key because it confirms that the solutions represented in the fundamental matrix form a basis for the solution space of the differential equation system. Hence, it enables the construction of the complete solution using initial conditions.

More practically, the test for linear independence within a matrix of solutions is accomplished using the Wronskian. If the Wronskian is non-zero (\(W eq 0\)), the solutions are linearly independent. This ensures that the solutions can span the entire space without one being redundant.

Inverse of a Matrix

The inverse of a matrix is pivotal in various applications, including solving systems of equations and transformation in linear algebra. In the context of differential equations, particularly those involving matrices, the inverse is leveraged to understand transformation relationships between solutions.

For a matrix \(Y\), if \(\det(Y) eq 0\), it has an inverse \(Y^{-1}\). This inverse matrix plays a vital role when exploring new systems of equations derived by changing the coefficients or transformations like taking the transpose. For example, in exploring the system \(\mathbf{y}' = -A^T(t) \mathbf{y}\), the inverse \(Y^{-1}\) acts as a fundamental matrix in this transformed system.

Understanding the inverse aids in re-expressing problems in simpler forms, proving relations, and confirming whether transformed expressions still satisfy fundamental conditions like linear independence across a changed set of equations.