Problem 16 - Differential Equations and Linear Algebra

Question

An \(n \times n\) matrix \(A\) that satisfies \(A^{k}=0\) for some \(k\) is called nilpotent. Show that the given matrix is nilpotent, and use Definition 7.4 .1 to determine \(e^{A t}\). $$A=\left[\begin{array}{lll} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]$$

Step-by-Step Solution

Answer

In this exercise, we found that the given matrix \(A\) is nilpotent with \(k = 3\) as \(A^3 = 0\). Using Definition 7.4.1, we then calculated the matrix exponential \(e^{At}\) to be: \(e^{At} = \begin{bmatrix} 1 & 0 & 0 \\ t & 1 & 0 \\ 0 & t & 1 - \dfrac{t^2}{2} \end{bmatrix}\)

Step 1: Show that the matrix A is nilpotent

First, let's find the powers of A until we find a zero matrix. \(A^2 = AA = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}\) \(A^3 = A^2A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0\) So, the value of k is 3 as we found that A^3 = 0. Hence, the given matrix A is nilpotent.

Step 2: Determine e^(At) using Definition 7.4.1

Now that we know that the matrix A is nilpotent with k = 3, we can calculate the matrix exponential using Definition 7.4.1: \(e^{At} = I + At + \dfrac{(At)^2}{2!} + \dfrac{(At)^3}{3!} + \cdots\) As A is nilpotent with k = 3, the terms for powers of At higher than 2 become zero. So we only need to compute the first 3 terms of the series: \(e^{At} = I + At + \dfrac{(At)^2}{2!}\) Let's first compute At and (At)^2. \(At = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} t = \begin{bmatrix} 0 & 0 & 0 \\ t & 0 & 0 \\ 0 & t & 0 \end{bmatrix}\) \((At)^2 = (At)(At) = \begin{bmatrix} 0 & 0 & 0 \\ t & 0 & 0 \\ 0 & t & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ t & 0 & 0 \\ 0 & t & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & t^2 & 0 \end{bmatrix}\) Now, let's compute e^(At): \(e^{At} = I + At + \dfrac{(At)^2}{2!} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ t & 0 & 0 \\ 0 & t & 0 \end{bmatrix} + \dfrac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & t^2 & 0 \end{bmatrix} \) \(e^{At} = \begin{bmatrix} 1 & 0 & 0 \\ t & 1 & 0 \\ 0 & t & 1 - \dfrac{t^2}{2} \end{bmatrix}\) So, the matrix exponential e^(At) is: \(e^{At} = \begin{bmatrix} 1 & 0 & 0 \\ t & 1 & 0 \\ 0 & t & 1 - \dfrac{t^2}{2} \end{bmatrix}\)

Key Concepts

Matrix Exponential

The matrix exponential, denoted as \(e^{At}\), is a crucial concept in various fields, including linear algebra and differential equations. It extends the idea of the exponential function from numbers to matrices. In practice, computing the exponential of a matrix \(A\) involves a power series expansion analogous to the scalar exponential:

• \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\)
• \(e^{A} = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots\)

Matrix exponential has practical applications, especially in solving systems of linear differential equations. When a matrix is nilpotent, like in our example where \(A^3 = 0\), the series terminates after a finite number of terms. This termination makes calculations much simpler. For the matrix \(A\) given in the problem, the exponential is found by computing:

• \(e^{At} = I + At + \frac{(At)^2}{2}\)

By substituting \(At\) and \((At)^2\), we obtain the matrix exponential, which greatly simplifies the problem solving.

Differential Equations

Differential equations involve equations that relate functions and their derivatives. They are used to model various phenomena in science and engineering, such as population dynamics, heat transfer, and fluid flow. The matrix exponential becomes particularly useful when dealing with linear differential equations involving matrices. For example, consider a system of first-order linear differential equations given by:

• \(\frac{d\mathbf{x}}{dt} = A \mathbf{x}\)

Here, \(\mathbf{x}\) is a vector and \(A\) is a matrix. The solution to this system is expressed using the matrix exponential:

• \(\mathbf{x}(t) = e^{At}\mathbf{x}(0)\)

This relation indicates that the state vector \(\mathbf{x}(t)\) at time \(t\) can be found by multiplying the initial state \(\mathbf{x}(0)\) by the matrix exponential \(e^{At}\). In cases where the matrix \(A\) is nilpotent, such as the exercise example, the exponential simplifies significantly, allowing for straightforward integration into solving differential equations.

Linear Algebra

Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between these spaces. It includes studying matrices, determinants, eigenvalues, and eigenvectors. In our discussion, understanding matrices is fundamental because they represent linear transformations.
A key concept in linear algebra is nilpotency. A matrix \( A \) is nilpotent if \(A^k = 0\) for some integer \(k\). This property means applying the linear transformation represented by \(A\) multiple times eventually results in zeroing out any vector. In the given example, \(A^3 = 0\), so \(A\) is nilpotent with index 3. This nilpotency property is critical when computing the matrix exponential since it implies the power series expansion for \(e^{At}\) terminates after a finite number of terms, simplifying calculations greatly. Understanding these concepts within linear algebra not only helps solve specific problems but also provides a foundation for tackling more complex mathematical topics and real-world applications.