Problem 11 - Differential Equations and Linear Algebra

Question

Consider the matrix \(A=\left[\begin{array}{ll}a & b \\ 0 & a\end{array}\right] .\) We can write \(A=B+C,\) where \(B=\left[\begin{array}{ll}a & 0 \\ 0 & a\end{array}\right]\) and \(C=\left[\begin{array}{ll}0 & b \\ 0 & 0\end{array}\right]\)(a) Verify that \(B C=C B\) (b) Verify that \(C^{2}=0_{2},\) and determine \(e^{C t}\) (c) Use property (1) of the matrix exponential function to find \(e^{A t}\)

Step-by-Step Solution

Answer

The matrix exponential \(e^{At}\) is given by: \(e^{At} = \left[\begin{array}{ll}e^{at} & bte^{at} \\\ 0 & e^{at}\end{array}\right]\)

Step 1: Verify that \(BC = CB\)

We calculate the products \(BC\) and \(CB\): - Calculate \(BC\): \(B=\left[\begin{array}{ll}a & 0 \\\ 0 & a\end{array}\right]\) and \(C=\left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]\) \(BC = \left[\begin{array}{ll}a & 0 \\\ 0 & a\end{array}\right]\left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & ab \\\ 0 & 0\end{array}\right]\) - Calculate \(CB\): \(C=\left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]\) and \(B=\left[\begin{array}{ll}a & 0 \\\ 0 & a\end{array}\right]\) \(CB = \left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]\left[\begin{array}{ll}a & 0 \\\ 0 & a\end{array}\right]=\left[\begin{array}{ll}0 & ab \\\ 0 & 0\end{array}\right]\) Since \(BC=CB=\left[\begin{array}{ll}0 & ab \\\ 0 & 0\end{array}\right]\), the condition \(BC=CB\) is verified.

Step 2: Verify that \(C^2 = 0_2\), and determine \(e^{Ct}\)

Calculate \(C^2\): \(C=\left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]\) \(C^2 = \left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & b \\\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\\ 0 & 0\end{array}\right]=0_2\) Now that we have verified \(C^2 = 0_2\), we can determine the matrix exponential \(e^{Ct}\): Recall that the exponential of a matrix \(M\) is defined as \(e^M = I + M + \frac{1}{2!} M^2 + \frac{1}{3!} M^3 + \cdots\) Since \(C^2=0_2\), we have \(C^n=0_2\) for \(n\geq 2\), and we can compute the exponential matrix as follows: \(e^{Ct} = I + Ct + \frac{1}{2!} (Ct)^2 + \frac{1}{3!} (Ct)^3 + \cdots = I + Ct\) Thus, \(e^{Ct} = \left[\begin{array}{ll}1 & bt \\\ 0 & 1\end{array}\right]\)

Step 3: Use property (1) of the matrix exponential function to find \(e^{At}\)

Property (1) of the matrix exponential function states that if \(AB = BA\), then \(e^{A + B} = e^A e^B\). We already verified that \(BC = CB\), so we can compute \(e^{At} = e^{(B + C)t}\) using property (1): \(e^{At} = e^{(B + C)t} = e^{Bt} e^{Ct}\) We already calculated \(e^{Ct}\) in step 2. Now we determine \(e^{Bt}\): \(B=\left[\begin{array}{ll}a & 0 \\\ 0 & a\end{array}\right]\) which can be written as \(B = aI\), where \(I\) is the identity matrix. Therefore, we have: \(e^{Bt} = e^{aIt} = \left[\begin{array}{ll}e^{at} & 0 \\\ 0 & e^{at}\end{array}\right]\) Now, we can compute \(e^{At}\): \(e^{At} = e^{Bt} e^{Ct} = \left[\begin{array}{ll}e^{at} & 0 \\\ 0 & e^{at}\end{array}\right] \left[\begin{array}{ll}1 & bt \\\ 0 & 1\end{array}\right] =\left[\begin{array}{ll}e^{at} & bte^{at} \\\ 0 & e^{at}\end{array}\right]\)

Key Concepts

Matrix Multiplication

Matrix multiplication is an essential operation in linear algebra, where two matrices are combined to form a new matrix. Each entry in the resulting matrix is a dot product of the corresponding row from the first matrix and the column from the second matrix. It's important to ensure that the matrices are compatible for multiplication, meaning the number of columns in the first matrix should be the same as the number of rows in the second.

• Compute element-wise: Take the dot product of rows and columns.
• Observe non-commutativity: Generally, for matrices A and B, \( AB eq BA \).
• Special cases include situations like \( AB = BA \) as when shown in this exercise.

Matrix multiplication is crucial for various applications, including solving system of equations, transformations in graphics, and many others.

Diagonal Matrices

A diagonal matrix is a special kind of square matrix where all entries outside the main diagonal are zero. The main diagonal can have non-zero elements. Diagonal matrices are straightforward when it comes to operations such as multiplication and exponentiation.

• Simplified operations: Multiplying diagonal matrices is as easy as multiplying their diagonal elements individually.
• Matrix exponentials: The exponential of a diagonal matrix is also a diagonal matrix.
• Eigenvalues: The entries of a diagonal matrix are its eigenvalues.

A matrix like \( B = \left[ \begin{array}{cc} a & 0 \ 0 & a \end{array}\right] \) is diagonal and showcases the power of simplifying calculations with diagonal structures.

Nilpotent Matrices

Nilpotent matrices are a fascinating concept in matrix algebra. A matrix is nilpotent if there exists a natural number \( k \) such that \( C^k = 0 \). In the given exercise, matrix \( C \) is shown to be nilpotent because \( C^2 = 0 \).

• Vanishing powers: For a nilpotent matrix \( C \), starting from a certain power, the matrix becomes a zero matrix.
• Simplifying exponentials: For \( e^{Ct} \), often it reduces to \( I + Ct \) due to the nilpotency (since higher powers vanish).
• Applications: Nilpotent matrices appear in the study of linear transformations, particularly in decompositions and solving differential equations.

Understanding nilpotent matrices can greatly simplify complex matrix operations and has profound implications in theoretical mathematics.