Question

Let \(\mathbf{J}=\left(\begin{array}{cc}{\lambda} & {1} \\ {0} & {\lambda}\end{array}\right),\) where \(\lambda\) is an arbitrary real number. (a) Find \(\mathbf{J}^{2}, \mathbf{J}^{3},\) and \(\mathbf{J}^{4}\) (b) Use an inductive argument to show that \(\mathbf{J}^{n}=\left(\begin{array}{cc}{\lambda^{n}} & {n \lambda^{n-1}} \\\ {0} & {\lambda^{n}}\end{array}\right)\) (c) Determine exp(Jt). (d) Use exp(Jt) to solve the initial value problem \(\mathbf{x}^{\prime}=\mathbf{J x}, \mathbf{x}(0)=\mathbf{x}^{0}\)

Short answer

Answer: The solution of the initial value problem is given by \(x(t) = \begin{pmatrix}(x_1^0 + tx_2^0)e^{\lambda t} \\ x_2^0e^{\lambda t}\end{pmatrix}\), where \(J = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix}\) and \(x^0 = \begin{pmatrix}x_1^0 \\ x_2^0\end{pmatrix}\).

Step-by-step solution

Calculate J^2, J^3, and J^4

Let's find J^2 by multiplying J by itself: \(\mathbf{J}^2 = \mathbf{J}\mathbf{J}\) Use matrix multiplication to get: \(\mathbf{J}^2 = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} = \begin{pmatrix}\lambda^2 & 2\lambda \\ 0 & \lambda^2 \end{pmatrix}\) Similarly, we can find J^3 and J^4: \(\mathbf{J}^3 = \mathbf{J}\mathbf{J}^2 = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda^2 & 2\lambda \\ 0 & \lambda^2 \end{pmatrix} = \begin{pmatrix}\lambda^3 & 3\lambda^2 \\ 0 & \lambda^3 \end{pmatrix}\) \(\mathbf{J}^4 = \mathbf{J}\mathbf{J}^3 = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda^3 & 3\lambda^2 \\ 0 & \lambda^3 \end{pmatrix} = \begin{pmatrix}\lambda^4 & 4\lambda^3 \\ 0 & \lambda^4 \end{pmatrix}\)

Inductive Argument for J^n

We will use induction to prove that \(\mathbf{J}^n=\left(\begin{array}{cc}{\lambda^{n}} & {n \lambda^{n-1}} \\ {0} & {\lambda^{n}}\end{array}\right)\) - Base Case (n=1): We see that J^1 is already given as \(\mathbf{J} =\left(\begin{array}{cc}{\lambda^1} & {1\lambda^0} \\ {0} & {\lambda^1}\end{array}\right)\) - Inductive Step: Assume that the formula is true for n=k, \(\mathbf{J}^k=\left(\begin{array}{cc}{\lambda^{k}} & {k \lambda^{k-1}} \\ {0} & {\lambda^{k}}\end{array}\right)\) We will show that the formula is true for n=k+1, \(\mathbf{J}^{k+1}=\mathbf{J}\mathbf{J}^k = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda^k & k\lambda^{k-1} \\ 0 & \lambda^k \end{pmatrix}\) Using matrix multiplication, we get \(\mathbf{J}^{k+1}=\left(\begin{array}{cc}{\lambda^{k+1}} & {(k+1) \lambda^{k}} \\ {0} & {\lambda^{k+1}}\end{array}\right)\) Thus, the formula holds for n=k+1, and by induction, it holds for all positive integers n.

Calculate exp(Jt)

To calculate the matrix exponential exp(Jt), we first recall the matrix exponential formula: \(e^{\mathbf{A}}=\sum_{n=0}^{\infty}\frac{1}{n!}\mathbf{A}^n\) Applying this to Jt: \(e^{\mathbf{Jt}}=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\mathbf{Jt}\right)^n\) We found the formula for J^n in step 2. So, we can write: \(e^{\mathbf{Jt}}=\sum_{n=0}^{\infty}\frac{1}{n!}\begin{pmatrix}(\lambda t)^n & n(\lambda t)^{n-1}t \\ 0 & (\lambda t)^n\end{pmatrix}\) Now, we sum the series component-wise: \(e^{\mathbf{Jt}}=\begin{pmatrix}\sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!} & \sum_{n=0}^{\infty}\frac{n(\lambda t)^{n-1}t}{n!} \\ 0 & \sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!}\end{pmatrix}\) Notice the elements in the diagonal are just the exponential function of λt: \(e^{\mathbf{Jt}}=\begin{pmatrix}e^{\lambda t} & \sum_{n=0}^{\infty}\frac{n(\lambda t)^{n-1}t}{n!} \\ 0 & e^{\lambda t}\end{pmatrix}\) The series in the top-right element can be shown to be te^{\lambda t} (the first term is 0): \(e^{\mathbf{Jt}}=\begin{pmatrix}e^{\lambda t} & te^{\lambda t} \\ 0 & e^{\lambda t}\end{pmatrix}\)

Solve the Initial Value Problem

Given \(\mathbf{x'}=\mathbf{Jx}\) and \(\mathbf{x}(0)=\mathbf{x}^0\), we can solve the initial value problem using the matrix exponential: \(\mathbf{x}(t) = e^{\mathbf{Jt}}\mathbf{x}^0\) Substitute the matrix exponential found in step 3: \(\mathbf{x}(t) = \begin{pmatrix}e^{\lambda t} & te^{\lambda t} \\ 0 & e^{\lambda t}\end{pmatrix}\mathbf{x}^0\) Suppose \(\mathbf{x}^0 = \begin{pmatrix}x_1^0 \\ x_2^0\end{pmatrix}\). Then the solution is: \(\mathbf{x}(t) = \begin{pmatrix}e^{\lambda t} & te^{\lambda t} \\ 0 & e^{\lambda t}\end{pmatrix}\begin{pmatrix}x_1^0 \\ x_2^0\end{pmatrix}= \begin{pmatrix}(x_1^0 + tx_2^0)e^{\lambda t} \\ x_2^0e^{\lambda t}\end{pmatrix}\) The solution of the initial value problem is given by: \(\mathbf{x}(t) = \begin{pmatrix}(x_1^0 + tx_2^0)e^{\lambda t} \\ x_2^0e^{\lambda t}\end{pmatrix}\)

Key concepts

Jordan Canonical Form

The Jordan canonical form is a powerful mathematical tool used in linear algebra to simplify complex matrix computations. Imagine you have a bunch of building blocks, and some of them are pretty tricky to work with. The Jordan canonical form turns these blocks into a standardized set, so they're easier to handle. It reorganizes a square matrix into a block diagonal matrix where each block is called a Jordan block.In our exercise, the matrix \[\begin{equation}\mathbf{J}=\left(\begin{array}{cc}{\lambda} & {1} \ {0} &{\lambda}\end{array}\right),\end{equation}\]where \(\lambda\) is a specific real number, is already in a form similar to a Jordan block. This form helps simplify the process of raising the matrix to higher powers. The off-diagonal entry of '1' is what ties this back to the concept of Jordan canonical form, which often includes such off-diagonal ones in its Jordan blocks.

Initial Value Problem

In the world of differential equations, an initial value problem is like a detective case: you have a dynamic situation described by the differential equation, and you're given a crucial clue—the initial condition—to figure out the full story. It sets the stage for the equation's behavior from a specific starting point, allowing you to predict the future states of the system.For the problem at hand, our initial condition is \(\mathbf{x}(0)=\mathbf{x}^0\). This little piece of information, when combined with the differential equation \(\mathbf{x'}=\mathbf{Jx}\), allows us to unravel the mystery of how \(\mathbf{x}(t)\) behaves as time moves forward. By tackling the initial value problem using matrix exponentiation, solving for \(\mathbf{x}(t)\) becomes manageable, instead of an intimidating quest.

Matrix Exponential

The matrix exponential, denoted by \(e^{\mathbf{A}}\), acts like a magical transformation: it turns a system of linear differential equations into a formula for future values based on the current state. If our matrix is a recipe book for how things change over time, then the matrix exponential tells us the end result of following those recipes for 't' amount of time—a critical concept for solving differential equations.In the step-by-step solution of the exercise, we applied the matrix exponential to matrix \(\mathbf{J}\), incorporating time \(t\), and discovered a pattern emerging as we expanded the exponential series. This led us to find the exponential of our Jordan matrix over time, \(e^{\mathbf{Jt}}\), which is key to solving our initial value problem.

Inductive Proof

An inductive proof is like climbing a ladder; if you show the first rung (base case) is sturdy and that if one rung is stable (inductive hypothesis), the next one will be too (inductive step), then you can trust the ladder will hold all the way up. It's a fundamental strategy in mathematics for proving statements about infinite sequences, and in this context, we used it to establish the pattern for higher powers of our matrix \(\mathbf{J}\).The ladder analogy comes alive as we first verify the base case—showing our claim holds for \(n=1\). Then, by assuming the formula works for some arbitrary step 'k', we prove that it must work for the next one, 'k+1'. This process, applied to matrix \(\mathbf{J}\), illuminates a clear pattern leading to the formula for \(\mathbf{J}^n\), which is instrumental for finding the matrix exponential.