Question

Consider the system $$ \mathbf{x}^{\prime}=\mathbf{A x}=\left(\begin{array}{rrr}{5} & {-3} & {-2} \\\ {8} & {-5} & {-2} \\ {-4} & {-5} & {-4} \\ {-4} & {3} & {3}\end{array}\right) \mathbf{x} $$ (a) Show that \(r=1\) is a triple eigenvalue of the coefficient matrix \(\mathbf{A},\) and that there are only two linearly independent eigenvectors, which we may take as $$ \xi^{(1)}=\left(\begin{array}{l}{1} \\ {0} \\ {2}\end{array}\right), \quad \xi^{(2)}=\left(\begin{array}{r}{0} \\ {2} \\ {-3}\end{array}\right) $$ Find two linearly independent solutions \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) of Eq. (i). (b) To find a third solution assume that \(\mathbf{x}=\xi t e^{t}+\mathbf{\eta} e^{\lambda} ;\) thow that \(\xi\) and \(\eta\) must satisfy $$ (\mathbf{A}-\mathbf{1}) \xi=0 $$ \((\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}\) (c) Show that \(\xi=c_{1} \xi^{(1)}+c_{2} \mathbf{\xi}^{(2)},\) where \(c_{1}\) and \(c_{2}\) are arbitrary constants, is the most general solution of Eq. (iii). Show that in order to solve Eq. (iv) it is necessary that \(c_{1}=c_{2}\) (d) It is convenient to choose \(c_{1}=c_{2}=2 .\) For this choice show that $$ \xi=\left(\begin{array}{r}{2} \\ {4} \\ {-2}\end{array}\right), \quad \mathbf{\eta}=\left(\begin{array}{r}{0} \\ {0} \\ {-1}\end{array}\right) $$ where we have dropped the multiples of \(\xi^{(1)}\) and \(\xi^{(2)}\) that appear in \(\eta\). Use the results given in Eqs. (v) to find a third linearly independent solution \(\mathbf{x}^{(3)}\) of Eq. (i). (e) Write down a fundamental matrix \(\Psi(t)\) for the system (i). (f) Form a matrix T with the cigenvector \(\xi^{(1)}\) in the first column and with the eigenvector \(\xi\) and the generalized eigenvector \(\eta\) from Eqs. (v) in the other two columns. Find \(\mathbf{T}^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A} .\)

Short answer

The system has a triple eigenvalue r=1 with linearly independent eigenvectors $\xi^{(1)}$ and $\xi^{(2)}$, resulting in solutions $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$. We assume a third solution form $\mathbf{x}=\xi t e^{t}+\eta e^{\lambda}$, and after solving for ξ and η, obtain the third linearly independent solution $\mathbf{x}^{(3)}(t)$. The fundamental matrix is formed using these linearly independent solutions, and we find the Jordan form of A by computing the product $\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}$. The Jordan form of the matrix A is: $$ \mathbf{J}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right) $$

Step-by-step solution

(a) Triple eigenvalue and eigenvectors

To find the eigenvalues, we have to solve the characteristic equation \((\mathbf{A} - r\mathbf{I})\mathbf{\xi}=0\). The triple eigenvalue r=1 is given. So, let's find the eigenvectors by solving \((\mathbf{A} - \mathbf{I})\mathbf{\xi}=0\), where \(\mathbf{I}\) is an identity matrix. $$ \left(\begin{array}{ccc}4 & -3 & -2 \\ 8 & -6 & -2 \\ -4 & -5 & -5 \\ -4 & 3 & 2\end{array}\right)\mathbf{\xi}=0 $$ The given eigenvectors are: $$ \xi^{(1)}=\left(\begin{array}{l}{1}\\\ {0} \\\ {2}\end{array}\right), \quad \xi^{(2)}=\left(\begin{array}{r}{0}\\\ {2} \\\ {-3}\end{array}\right) $$ We can verify that \((\mathbf{A} - \mathbf{I})\mathbf{\xi}^{(1)}=0\) and \((\mathbf{A} - \mathbf{I})\mathbf{\xi}^{(2)}=0\), which confirms they are eigenvectors for the eigenvalue \(r=1\). The solutions are: $$ \mathbf{x}^{(1)}(t)=\mathbf{\xi}^{(1)}e^{t}=\left(\begin{array}{l}{1}\\\ {0} \\\ {2}\end{array}\right)e^{t}, \quad \mathbf{x}^{(2)}(t)=\mathbf{\xi}^{(2)}e^{t}=\left(\begin{array}{r}{0}\\\ {2} \\\ {-3}\end{array}\right)e^{t} $$

(b) Third solution

Let's assume a third solution: $$ \mathbf{x}=\xi t e^{t}+\mathbf{\eta}e^{\lambda} ; $$ where ξ and η need to satisfy: $$ (\mathbf{A}-\mathbf{1}) \xi=0, \quad (\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}$$

(c) General solution for ξ

Now, let's write the most general solution for ξ: $$ \xi=c_{1} \xi^{(1)}+c_{2} \mathbf{\xi}^{(2)} $$ In order to solve \((\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}\), it is necessary that \(c_{1}=c_{2}\).

(d) Specific choice of constants

Let's choose \(c_{1}=c_{2}=2\): $$ \xi=\left(\begin{array}{r}{2} \\\ {4} \\\ {-2}\end{array}\right), \quad \mathbf{\eta}=\left(\begin{array}{r}{0} \\\ {0} \\\ {-1}\end{array}\right) $$ The third linearly independent solution is given by: $$ \mathbf{x}^{(3)}(t)=\xi te^{t}+\mathbf{\eta}e^{t}=\left(\begin{array}{r}{2t} \\\ {4t} \\\ {-2t-1}\end{array}\right)e^{t} $$

(e) Fundamental matrix

The fundamental matrix \(\Psi(t)\) can be formed using the linearly independent solutions: $$ \Psi(t) = \left(\begin{array}{ccc} e^{t} & 0 & 2te^{t} \\ 0 & 2e^{t} & 4te^{t} \\ 2e^{t} & -3e^{t} & (-2t-1)e^{t} \end{array}\right) $$

(f) Jordan form of A

Now we need to form a matrix T with the eigenvector \(\xi^{(1)}\) in the first column, and the eigenvector \(\xi\) and the generalized eigenvector \(\eta\) from Eqs. (v) in the other two columns: $$ \mathbf{T}=\left(\begin{array}{ccc}1 & 2 & 0 \\ 0 & 4 & 0 \\ 2 & -2 & -1\end{array}\right) $$ Calculating the inverse of T: $$ \mathbf{T}^{-1}=\left(\begin{array}{ccc}1 & -1 & 0 \\ 0 & 1/2 & 0 \\ -4 & 10 & -1\end{array}\right) $$ Finally, let's compute the product to obtain the Jordan form of A: $$ \mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right) $$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts in linear algebra, playing a critical role in various applications including the study of differential equations. An eigenvalue is a scalar that indicates how much a corresponding eigenvector is stretched or compressed during a linear transformation represented by a matrix.

To find the eigenvalues of a matrix, one must solve the characteristic equation \(\text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{A}\) is the matrix in question, \(\lambda\) represents the eigenvalue, and \(\mathbf{I}\) is the identity matrix. The corresponding eigenvectors are non-zero vectors that satisfy the equation \(\mathbf{A}\mathbf{v} = \lambda \mathbf{v}\).

In the context of the exercise provided, identifying the appropriate eigenvalues and eigenvectors is essential to solving the differential equation, as illustrated in Step 1 of the solution. These mathematical objects reveal the behavior of the system's solutions over time and are used to construct independent solutions to the system of equations.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. Its concepts are not just theoretical; they are extremely useful in practical applications ranging from natural sciences to social sciences, emphasizing its fundamental role in the system of differential equations under discussion.

For instance, finding a set of linearly independent eigenvectors and using them to determine solutions to the system falls within the purview of linear algebra. However, as observed in Step 1 of the solution, there might be cases when we cannot find a complete set of linearly independent eigenvectors to span the solution space. In such scenarios, generalized eigenvectors and the concept of the Jordan form come into play to find a comprehensive solution.

Matrix Exponential

The matrix exponential is a significant concept in solving systems of linear differential equations. Defined analogously to the exponential function for real numbers, the matrix exponential provides a way to solve linear systems of differential equations where the solution evolves over time.

In essence, applying the matrix exponential to a matrix yields a new matrix that can express the evolution of a system described by a set of linear differential equations. In Step 5 of the solution, the fundamental matrix \(\Psi(t)\) includes terms that are essentially the matrix exponential of the coefficient matrix times \(t\), represented by the notation \(e^{\mathbf{A}t}\). It embodies all the solutions to the system and is key in understanding the system's dynamics.

Jordan Form

The Jordan form, or Jordan canonical form, is a simplified matrix that is similar to a given square matrix, demonstrating the same geometric features. Particularly useful when a matrix does not have enough linearly independent eigenvectors to diagonalize, it includes both the eigenvalues and the structure of the matrix related to its eigenspaces.

In simpler terms, the Jordan form is a block diagonal matrix with each block corresponding to an eigenvalue and possibly having ones on the super diagonal if generalized eigenvectors are involved. This form simplifies the power of the matrix, making it easier to compute functions of the matrix such as the matrix exponential.

The exercise improvement advice emphasizes the importance of understanding the process leading to the Jordan form, as seen in Step 6 of the solution, where the Jordan form of matrix \(\mathbf{A}\) is obtained. Grasping the Jordan form allows students to better analyze and solve systems that can't be diagonalized in the usual sense, highlighting the direct application of linear algebra to differential equations.