Chapter 4: Problem 4
Find the stable, unstable and center subspaces, \(E^{*}, E^{u}\) and \(E^{c}\) for the linear maps \(L(x)=A x\) where the matrix (a) \(A=\left[\begin{array}{rr}2 & 0 \\ 1 & -1\end{array}\right]\) (b) \(A=\left[\begin{array}{cc}1 / 2 & 1 \\ 0 & 1 / 2\end{array}\right]\) (c) \(A=\left[\begin{array}{rr}1 & -1 \\ 1 & 1\end{array}\right]\) (d) \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]\).
Short Answer
Step by step solution
- Find the Eigenvalues
- Find the Eigenvectors
- Classify the Subspaces
- Subspace Calculation for Part (a)
- Subspace Calculation for Part (b)
- Subspace Calculation for Part (c)
- Subspace Calculation for Part (d)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
stable subspace
For example, consider matrix (a) from the exercise: \(A = \begin{pmatrix} 2 & 0 \ 1 & -1 \end{pmatrix}\). The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -1\). Since \(\lambda_2 = -1\) has a negative real part, the stable subspace is spanned by the eigenvector corresponding to \(\lambda_2\), which is \(\begin{pmatrix} 0 \ 1 \end{pmatrix}\). Thus, \(E^*\) is \(\text{Span}\{\begin{pmatrix} 0 \ 1 \end{pmatrix}\}\).
Understanding stable subspaces helps in analyzing the long-term behavior of systems and predicting which components will diminish over time.
unstable subspace
For instance, considering matrix (d) from the exercise: \(A = \begin{pmatrix} 1 & 1 \ 1 & 2 \end{pmatrix}\), the eigenvalues are \(\lambda_1 = 1 + \sqrt{2}\) and \(\lambda_2 = 1 - \sqrt{2}\). Since \(\lambda_1 = 1 + \sqrt{2}\) has a positive real part, the unstable subspace is spanned by the eigenvector corresponding to \(\lambda_1\), which is \(\begin{pmatrix} 1 \ 1 + \sqrt{2} \end{pmatrix}\). Thus, \(E^u\) is \(\text{Span}\{\begin{pmatrix} 1 \ 1 + \sqrt{2} \end{pmatrix}\}\).
Identifying the unstable subspace is crucial in understanding how a system can move away from equilibrium and potentially cause system divergence.
center subspace
For example, consider matrix (b) from the exercise: \(A = \begin{pmatrix} \frac{1}{2} & 1 \ 0 & \frac{1}{2} \end{pmatrix}\). The eigenvalues are \(\lambda_1 = \frac{1}{2}\) and \(\lambda_2 = \frac{1}{2}\), having zero real parts, resulting in a center subspace spanned by the eigenvectors \(\begin{pmatrix} 1 \ 0 \end{pmatrix}\) and \(\begin{pmatrix} \frac{1}{2} \ 1 \end{pmatrix}\). Thus, \(E^c\) is \(\text{Span}\{\begin{pmatrix} 1 \ 0 \end{pmatrix}, \begin{pmatrix} \frac{1}{2} \ 1 \end{pmatrix}\}\).
The center subspace reveals crucial information about the components of a system that remain constant and do not change over time.
linear maps
Linear maps possess properties such as additivity and homogeneity, meaning \(L(x + y) = L(x) + L(y)\) and \(L(cx) = cL(x)\) for any vectors \(x\) and \(y\), and any scalar \(c\). These maps play a fundamental role in linear algebra and are essential in analyzing the behavior of dynamic systems.
For example, with matrix (c) from the exercise: \(A = \begin{pmatrix} 1 & -1 \ 1 & 1 \end{pmatrix}\), the application of the linear map to a vector involves multiplying \(A\) by the vector \(x\). Linear maps provide a straightforward method to understand complex transformations.
matrix analysis
In this exercise, matrix analysis was used to compute the eigenvalues and eigenvectors of given matrices. For instance, for matrix (a): \(A = \begin{pmatrix} 2 & 0 \ 1 & -1 \end{pmatrix}\), the characteristic polynomial \(\det(A - \lambda I) = 0\) yields eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = -1\). The corresponding eigenvectors were then used to classify the stable and unstable subspaces.
Mastering matrix analysis is critical for understanding linear transformations and solving systems of linear equations.