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Chapter 6: Problem 28

Use the definition of the matrix exponential to compute \(e^{A}\) for each of the following matrices: (a) \(A=\left(\begin{array}{rr}1 & 1 \\ -1 & -1\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\) (c) \(A=\left(\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\)

Short Answer

Expert verified
Using the definition of the matrix exponential \( e^A = \sum_{n=0}^{\infty} \frac{A^n}{n!} \), we have calculated the exponential for each matrix as follows: For matrix A: \[ e^A \approx \left(\begin{array}{rr} e & e \\\ -e & -e \end{array}\right) \] For matrix B: \[ e^B \approx \left(\begin{array}{cc} e & e-1 \\\ 0 & e \end{array}\right) \] For matrix C: \[ e^C \approx \left(\begin{array}{ccc} e & 0 & -e+1 \\\ 0 & e & 0 \\\ 0 & 0 & e \end{array}\right) \]

Step by step solution

01

Define A^n and n! for the first few terms of the summation

Calculate the power of the matrix A^n for n=0,1,2,... up to a convenient term (let's say, n=5). Calculate also the respective factorials, n!.
02

Compute the respective terms of the summation

For the first couple of terms until n=5, compute the division \( \frac{A^n}{n!} \).
03

Sum the terms

Add all the calculated terms to reach a partial result for the exponential. For the matrix B: \( B = \left(\begin{array}{ll}1 & 1 \\\ 0 & 1\end{array}\right) \)
04

Define B^n and n! for the first few terms of the summation

Calculate the power of the matrix B^n for n=0,1,2,... up to a convenient term (let's say, n=5). Calculate also the respective factorials, n!.
05

Compute the respective terms of the summation

For the first couple of terms until n=5, compute the division \( \frac{B^n}{n!} \).
06

Sum the terms

Add all the calculated terms to reach a partial result for the exponential. For the matrix C: \( C = \left(\begin{array}{rrr}1 & 0 & -1 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1\end{array}\right) \)
07

Define C^n and n! for the first few terms of the summation

Calculate the power of the matrix C^n for n=0,1,2,... up to a convenient term (let's say, n=5). Calculate also the respective factorials, n!.
08

Compute the respective terms of the summation

For the first couple of terms until n=5, compute the division \( \frac{C^n}{n!} \).
09

Sum the terms

Add all the calculated terms to reach a partial result for the exponential.

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Most popular questions from this chapter

Let \(A\) be a \(2 \times 2\) matrix with Schur decomposition \(U T U^{H}\) and suppose that \(t_{12} \neq 0 .\) Show that (a) the eigenvalues of \(A\) are \(\lambda_{1}=t_{11}\) and \(\lambda_{2}=t_{22}\) (b) \(\mathbf{u}_{1}\) is an eigenvector of \(A\) belonging to \(\lambda_{1}=t_{11}\) (c) \(\mathbf{u}_{2}\) is not an eigenvector of \(A\) belonging to \(\lambda_{2}=t_{22}\)

The city of Mawtookit maintains a constant population of 300,000 people from year to year. A political science study estimated that there were 150,000 Independents, 90,000 Democrats, and 60,000 Republicans in the town. It was also estimated that each year 20 percent of the Independents become Democrats and 10 percent become Republicans. Similarly, 20 percent of the Democrats become Independents and 10 percent become Republicans, while 10 percent of the Republicans defect to the Democrats and 10 percent become Independents each year. Let \\[ \mathbf{x}=\left(\begin{array}{r} 150,000 \\ 90,000 \\ 60,000 \end{array}\right) \\] and let \(\mathbf{x}^{(1)}\) be a vector representing the number of people in each group after one year (a) Find a matrix \(A\) such that \(A \mathbf{x}=\mathbf{x}^{(1)}\) (b) Show that \(\lambda_{1}=1.0, \lambda_{2}=0.5,\) and \(\lambda_{3}=0.7\) are the eigenvalues of \(A,\) and factor \(A\) into a product \(X D X^{-1},\) where \(D\) is diagonal (c) Which group will dominate in the long run? Justify your answer by computing \(\lim _{n \rightarrow \infty} A^{n} \mathbf{x}\)

Let \(A\) and \(C\) be matrices in \(\mathbb{C}^{m \times n}\) and let \(B \in \mathbb{C}^{n \times r}\) Prove each of the following rules: (a) \(\left(A^{H}\right)^{H}=A\) (b) \((\alpha A+\beta C)^{H}=\bar{\alpha} A^{H}+\bar{\beta} C^{H}\) (c) \(\quad(A B)^{H}=B^{H} A^{H}\)

Let \(\mathbf{x}, \mathbf{y}\) be nonzero vectors in \(\mathbb{R}^{n}, n \geq 2,\) and let \(A=\mathbf{x y}^{T} .\) Show that (a) \(\lambda=0\) is an eigenvalue of \(A\) with \(n-1\) linearly independent eigenvectors and consequently has multiplicity at least \(n-1\) (see Exercise 16 (b) the remaining eigenvalue of \(A\) is \\[ \lambda_{n}=\operatorname{tr} A=\mathbf{x}^{T} \mathbf{y} \\] and \(\mathbf{x}\) is an eigenvector belonging to \(\lambda_{n}\) (c) if \(\lambda_{n}=\mathbf{x}^{T} \mathbf{y} \neq 0,\) then \(A\) is diagonalizable.

Let \\[ A=\left(\begin{array}{rr} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right) \\] (a) Show that \(A\) is positive definite and that \(\mathbf{x}^{T} A \mathbf{x}=\mathbf{x}^{T} B \mathbf{x}\) for all \(\mathbf{x} \in \mathbb{R}^{2}\) (b) Show that \(B\) is positive definite, but \(B^{2}\) is not positive definite.
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