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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 \(\mathbf{n}_{0}=\left[ \begin{array}{l}{1} \\ {1}\end{array}\right]\). \(A=\left[ \begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]\)

Short Answer

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

Step by step solution

01

Find the Eigenvalues of Matrix A

The given matrix is \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \). To find the eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \). This means we solve \( \text{det}\begin{bmatrix} 0 - \lambda & 1 \ 1 & 0 - \lambda \end{bmatrix} = 0 \), leading to \( \lambda^2 - 1 = 0 \). The solutions are \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \).
02

Find the Eigenvectors of Matrix A

With eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \), find the corresponding eigenvectors. For \( \lambda_1 = 1 \): Solve \( (A - I)\mathbf{v} = 0 \). This gives the system \( -v_1 + v_2 = 0 \), leading to eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \). For \( \lambda_2 = -1 \): Solve \( (A + I)\mathbf{v} = 0 \). This gives the system \( v_1 + v_2 = 0 \), leading to eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \).
03

Express Initial Vector in Terms of Eigenvectors

The initial vector is \( \mathbf{n}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \), which is actually the first eigenvector \( \mathbf{v}_1 \). Hence, \( \mathbf{n}_0 \) can be expressed as \( \mathbf{n}_0 = 1 \cdot \mathbf{v}_1 \).
04

Use the Recursive Relationship to Express \( \mathbf{n}_t \)

The solution to the recursion \( \mathbf{n}_{t+1} = A \mathbf{n}_t \) uses \( \mathbf{n}_t = c_1 \lambda_1^t \mathbf{v}_1 + c_2 \lambda_2^t \mathbf{v}_2 \). Since \( \mathbf{n}_0 = \mathbf{v}_1 \), we have \( c_1 = 1 \) and \( c_2 = 0 \). Thus, \( \mathbf{n}_t = \lambda_1^t \mathbf{v}_1 \).
05

Solve and Simplify Expression for \( \mathbf{n}_t \)

With \( \lambda_1 = 1 \), the expression simplifies to \( \mathbf{n}_t = 1^t \cdot \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \). Therefore, each iteration of the recursion provides the same vector.

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Key Concepts

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

Eigenvalues
Eigenvalues are special numbers associated with a matrix that give us a sense of its behavior in certain transformations. For a given square matrix, eigenvalues can be thought of as scalars that stretch or shrink vectors in a particular direction during the transformation.

Calculating eigenvalues involves setting up the characteristic equation, which is derived by subtracting a scalar multiple of the identity matrix, often denoted as \(I\), from the original matrix \(A\). This results in \(A - \lambda I\), where \(\lambda\) represents the eigenvalues. By calculating the determinant \(\text{det}(A - \lambda I)\) and setting it to zero, we find the possible values of \(\lambda\) that satisfy this equation.

In our example, the matrix \(A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}\) leads to the characteristic equation \(\text{det}(A - \lambda I) = 0\). Solving this equation reveals the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = -1\). These indicate how the matrix affects vectors in its transformation.
Eigenvectors
Once we have the eigenvalues, we can find the corresponding eigenvectors. Eigenvectors are non-zero vectors that, when transformed by a matrix, result in a vector that is a scaled version of the original.

To find an eigenvector associated with a specific eigenvalue \(\lambda\), we solve the equation \((A - \lambda I)\mathbf{v} = 0\). The solutions are the eigenvectors \(\mathbf{v}\) corresponding to the matrix \(A\).

In the case of our matrix \(A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}\), the eigenvector associated with \(\lambda_1 = 1\) is \(\mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\). For \(\lambda_2 = -1\), the eigenvector is \(\mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix}\). These eigenvectors reflect the directions in which the matrix transforms the vectors.
Matrix Algebra
Matrix algebra involves operations such as addition, multiplication, and finding determinants, which are fundamental in analyzing and solving systems represented by matrices.

It plays a crucial role in linear transformations, allowing us to understand and express complex systems succinctly. In our discussion, we used matrix algebra to derive the characteristic equation and compute eigenvalues and eigenvectors.

Understanding these operations lets us model transformations and changes in recursive systems. For instance, when we expressed a recursion in terms of eigenvectors, we represented the system behavior over time using matrix multiplication, depicting how future states evolve from the initial inputs and transformations prescribed by \(A\).
Recursion Formula
Recursion involves expressing each term of a sequence based on its predecessors, often useful in dynamic systems and computer algorithms.

In linear algebra, recursion can describe how a system evolves over discrete steps, making it vital for understanding iterative processes. For the case of \(\mathbf{n}_{t+1} = A\mathbf{n}_t\), the recursion formula helps determine subsequent vectors \(\mathbf{n}_t\), given an initial vector \(\mathbf{n}_0\).

In our example, starting from \(\mathbf{n}_0 = \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\), we find each \(\mathbf{n}_t\) using its factorized form: \(\mathbf{n}_t = c_1 \lambda_1^t \mathbf{v}_1 + c_2 \lambda_2^t \mathbf{v}_2\). Given the initial conditions, this simplifies to \(\mathbf{n}_t = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) for all \(t\), since \(c_1 = 1\) and \(c_2 = 0\). This shows that the system reaches a steady state, maintaining its form through each recursion step.

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Most popular questions from this chapter

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

If \(\mathbf{a}=[3,0,-1],\) find a vector b such that \(\operatorname{comp}_{a} \mathbf{b}=2.\)

Describe and sketch a solid with the following properties: When illuminated by rays parallel to the \(z\) -axis, its shadow is a circular disk. If the rays are parallel to the y-axis, its shadow is a square. If the rays are parallel to the \(x\) -axis, its shadow is an isosceles triangle.

The case of matrices with repeated eigenvalues is treated in courses on linear algebra. As an example, try calculating the eigenvalues and eigenvectors of the following matrices. Comment on anything unusual that occurs. $${ (a) }\left[ \begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right] \quad(\mathrm{b}) \left[ \begin{array}{ll}{1} & {1} \\ {0} & {1}\end{array}\right] \quad \text { (c) } \left[ \begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {2}\end{array}\right]$$

Find the angle between a diagonal of a cube and a diagonal of one of its faces.

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