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Chapter 8: Problem 18

Express the solution to the recursion \(\mathbf{n}_{t+1}=A \mathbf{n}_{t}\) in terms of the eigenvectors and eigenvalues of \(A,\) assuming that \(\mathbf{n}_{0}=\left[ \begin{array}{l}{1} \\ {1}\end{array}\right]\). \(A=\left[ \begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]\)

Short Answer

Expert verified
The solution is \(\mathbf{n}_t = 2^t \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Step by step solution

01

Find Eigenvalues of Matrix A

To find the eigenvalues of matrix \(A\), we solve the characteristic equation \(\det(\lambda I - A) = 0\). For matrix \(A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\), the characteristic polynomial becomes \(\det\begin{bmatrix} \lambda - 1 & -1 \ -1 & \lambda - 1 \end{bmatrix} = 0\). Solving \((\lambda - 1)^2 - 1 = 0\) gives the eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = 0\).
02

Find Eigenvectors Corresponding to Each Eigenvalue

First, for \(\lambda_1 = 2\), solve \((A - 2I)\mathbf{v} = 0\). The resulting system \(-\mathbf{v}_1 + \mathbf{v}_2 = 0\) gives the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\).Next, for \(\lambda_2 = 0\), solve \((A - 0I)\mathbf{v} = 0\). The resulting system \(\mathbf{v}_1 + \mathbf{v}_2 = 0\) gives the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix}\).
03

Express Initial Vector as Linear Combination of Eigenvectors

Express \(\mathbf{n}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) as a combination of eigenvectors. Our initial condition \(\mathbf{n}_0 = 1\cdot\mathbf{v}_1 + 0\cdot\mathbf{v}_2\), since \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) is (1)(\(\begin{bmatrix} 1 \ 1 \end{bmatrix}\)). This implies \(c_1 = 1\) and \(c_2 = 0\).
04

Use Eigenvalues to Solve the Recursion

The solution \(\mathbf{n}_t\) is given by \(c_1\lambda_1^t \mathbf{v}_1 + c_2\lambda_2^t \mathbf{v}_2\). Substituting the values gives \(\mathbf{n}_t = 1(2^t)\begin{bmatrix} 1 \ 1 \end{bmatrix} + 0(0^t)\begin{bmatrix} 1 \ -1 \end{bmatrix}\).Thus, the expression simplifies to \(\mathbf{n}_t = 2^t\begin{bmatrix} 1 \ 1 \end{bmatrix}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recursion in Linear Algebra
In the world of linear algebra, recursion describes processes where each step depends on the previous one. This concept is similar to how some sequences work, based on previous terms. In our exercise, we examine a recursion formula:
\(\mathbf{n}_{t+1} = A \mathbf{n}_{t}\).
This equation describes how a vector, \(\mathbf{n}\), evolves over time, when multiplied by a matrix, \(A\).
  • Here, \(A\) represents a consistent transformation applied to \(\mathbf{n}\) during each iteration.
  • The matrix \(A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\) implements the transformation.
  • The initial vector, \(\mathbf{n}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\), serves as the starting point.
The solution to this recursion, therefore, depends on the matrix \(A\) and how it continuously transforms \(\mathbf{n}\). Our goal is to determine how this vector changes with each recursive step.
Characteristic Equation
The characteristic equation is a vital concept in discovering eigenvalues of a matrix. It arises when we set the determinant of \(\lambda I - A\) equal to zero, where \(I\) is the identity matrix, and \(\lambda\) are scalar values called eigenvalues. Let's break down the steps:
  • Use \(A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\) to form \(\lambda I - A\), resulting in \(\begin{bmatrix} \lambda - 1 & -1 \ -1 & \lambda - 1 \end{bmatrix}\).
  • Calculate the determinant and solve: \(\det(\lambda I - A) = (\lambda - 1)^2 - 1 = 0\).
  • This gives two solutions or eigenvalues: \(\lambda_1 = 2\) and \(\lambda_2 = 0\).
Eigenvalues are crucial because they provide insights into the matrix's behavior, particularly how vectors transform when applying \(A\). Finding these allows us to explore further into how this affects our recursion problem, giving fundaments to solve it.
Linear Combination of Vectors
Representing vectors as linear combinations adds the power to express complex transformations simply. The idea is to break a vector into the basic "directions" or fundamental vectors.
For our exercise:
  • The initial vector \(\mathbf{n}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) must be expressed in terms of eigenvectors.
  • Eigenvectors for \(\lambda_1 = 2\) and \(\lambda_2 = 0\) are \(\mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix}\), respectively.
  • Upon inspection, \(\mathbf{n}_0\) is already \(1\cdot\mathbf{v}_1 + 0\cdot\mathbf{v}_2\).
This format allows the component of the vector affected by \(\lambda_1\) to grow at a rate determined by \(\lambda_1^t\). Similarly, any contribution from \(\lambda_2\) would influence this if \(c_2\) were non-zero.
Ultimately, expressing vectors this way facilitates solving recursive equations precisely, projecting the initial conditions through the eigenvalues' growth.

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Most popular questions from this chapter

Fibonacci numbers The \(n\) th Fibonacci number is generated by the recursion equation \(F_{n}=F_{n-1}+F_{n-2}\) with \(F_{0}=F_{1}=1 .\) Define the two new variables \(x_{n}=F_{n}\) and \(y_{n}=F_{n-1} .\) (a) Show that the vector \(\mathbf{Z}_{n}=\left[ \begin{array}{l}{x_{n}} \\\ {y_{n}}\end{array}\right]\) obeys the recursion \(\mathbf{z}_{n}=A \mathbf{z}_{n-1},\) where $$A=\left[ \begin{array}{cc}{1} & {1} \\ {1} & {0}\end{array}\right]$$ (b) Find the eigenvalues of \(A\) (c) Can the Perron-Frobenius Theorem be applied to the matrix \(A ?\) If so, what is the long-term behavior of the vector \(\mathbf{Z}_{n} ?\) (d) Find a formula for the \(n\) th Fibonacci number.

$$\begin{array}{l}{\text { Biomechanics Two sprinters of equal mass leave the }} \\ {\text { starting blocks with the following horizontal and vertical }} \\\ {\text { force vectors: }}\end{array}$$ $$\begin{array}{|c|c|c|}\hline \text { Runner } & {\text { Horizontal }} & {\text { Vertical }} \\ \hline \text { Runner 1 } & {150 \mathrm{N}} & {300 \mathrm{N}} \\ \hline \text { Runner } 2 & {200 \mathrm{N}} & {250 \mathrm{N}} \\ \hline\end{array}$$ $$\begin{array}{l}{\text { Newton's Second Law states that force is equal to mass }} \\ {\text { times acceleration }(\mathbf{F}=m \mathbf{a}) . \text { Which runner has the greater }} \\ {\text { acceleration out of the blocks? }}\end{array}$$

\(\begin{array}{l}{\text { Find a second-degree polynomial (that is, an equation of }} \\ {\text { the form } y=a+b x+c x^{2} ) \text { that goes through the three }} \\ {\text { points }(0,1),(1,0), \text { and }(-1,0) . \text { Is there more than one }} \\ {\text { possibility? }}\end{array}\)

Express the solution to the recursion \(\mathbf{n}_{t+1}=A \mathbf{n}_{t}\) in terms of the eigenvectors and eigenvalues of \(A .\) Use DeMoivre's Theorem to simplify the solution if appropriate. \(A=\left[ \begin{array}{rr}{1} & {5} \\ {-1} & {1}\end{array}\right] \quad\) with \(\mathbf{n}_{0}=\left[ \begin{array}{l}{0} \\ {1}\end{array}\right]\)

Express the solution to the recursion \(\mathbf{n}_{t+1}=A \mathbf{n}_{t}\) in terms of the eigenvectors and eigenvalues of \(A .\) Use DeMoivre's Theorem to simplify the solution if appropriate. \(A=\left[ \begin{array}{rr}{1} & {-1} \\ {1} & {1}\end{array}\right] \quad\) with \(\mathbf{n}_{0}=\left[ \begin{array}{l}{1} \\ {1}\end{array}\right]\)
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