Chapter 5: Problem 18
Define \(T : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) by \(T(\mathbf{x})=A \mathbf{x},\) where \(A\) is a \(3 \times 3\) matrix with eigenvalues 5 and \(-2 .\) Does there exist a basis \(\mathcal{B}\) for \(\mathbb{R}^{3}\) such that the \(\mathcal{B}\) -matrix for \(T\) is a diagonal matrix? Discuss.
Short Answer
Expert verified
Yes, there exists a basis \( \mathcal{B} \) for \( \mathbb{R}^3 \) such that the \( \mathcal{B} \)-matrix for \( T \) is diagonal.
Step by step solution
01
Determine the Properties of Matrix A
Matrix A is a 3x3 matrix with eigenvalues 5 and -2. Since A is a square matrix, it can have up to 3 eigenvalues. However, given two eigenvalues, we need an additional eigenvalue to completely characterize A. For A to be diagonalizable, it must have 3 linearly independent eigenvectors. Including the necessary eigenvalues being repeated if not given explicitly.
02
Calculate Eigenvalue Multiplicities
A 3x3 matrix with eigenvalues 5 and -2, must have a total sum of eigenvalue multiplicities equal to 3. Since there are two specified eigenvalues, one possibility is that the eigenvalue 5 could have a multiplicity of 2 and -2 could have a multiplicity of 1, or vice versa.
03
Check for Diagonalizability
For A to be diagonalizable, it must have 3 linearly independent eigenvectors. This condition is met if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity. If the algebraic multiplicity adds up to 3 with 3 independent eigenvectors, it is diagonalizable.
04
Construct Potential Scenarios
Scenario 1: Eigenvalue 5 has an algebraic multiplicity of 2 and eigenvalue -2 has a multiplicity of 1. Scenario 2: Eigenvalue 5 has a multiplicity of 1 and -2 has a multiplicity of 2. In both scenarios, having each eigenvalue supported by its number of linearly independent eigenvectors makes the matrix A diagonalizable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues
In linear algebra, eigenvalues are crucial in understanding the characteristics of a square matrix like A. When a matrix is transformed, eigenvalues indicate how much a system "stretches" or "compresses" along its basis vectors. These values are found by solving the characteristic equation:
- \ (A - \lambda I) = 0 \ where \ \(\lambda\) represents the eigenvalues.
Eigenvectors
Eigenvectors are vectors associated with a particular eigenvalue that define directions preserved by a linear transformation. For a matrix transformation defined by T(x) = Ax, when A acts on its eigenvector, it scales the vector by the corresponding eigenvalue,
- i.e., \ A\mathbf{v} = \lambda\mathbf{v}, \ where \ \(\mathbf{v}\) is the eigenvector.
Algebraic Multiplicity
Algebraic multiplicity refers to the number of times an eigenvalue appears in the characteristic polynomial of a matrix. It indicates how many solutions exist for a certain eigenvalue when solving the characteristic equation. Since we are dealing with a 3x3 matrix A, the sum of the algebraic multiplicities of all eigenvalues must equal 3.
For example:
- If eigenvalue 5 has an algebraic multiplicity of 2 and eigenvalue -2 has a multiplicity of 1, it implies the eigenvalue 5 appears twice in the polynomial.
- Alternatively, if 5 has a multiplicity of 1 and -2 has a multiplicity of 2, these roles are reversed.
Geometric Multiplicity
The geometric multiplicity of an eigenvalue refers to the number of linearly independent eigenvectors corresponding to it. Every eigenvalue of a matrix has an associated eigenspace, and the dimension of this space is the geometric multiplicity. A fundamental condition for diagonalizability is that each eigenvalue’s geometric multiplicity must equal its algebraic multiplicity.
For our 3x3 matrix A:
- If eigenvalue 5 has a geometric multiplicity of 2, we should find 2 linearly independent eigenvectors associated with it.
- Similarly, if -2 has a geometric multiplicity of 1, then only one independent eigenvector is expected.