Math Problem Statement
Find the minimal polynomial associated with the matrix \(\begin{pmatrix} 0 & 0 & 3 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{pmatrix}\).
Solution
To determine the minimal polynomial of the matrix given in the image, we need to perform the following steps:
The matrix is:
0 & 0 & 3 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{pmatrix}$$ ### Step 1: Find the characteristic polynomial The characteristic polynomial of a matrix $$A$$ is given by: $$p_A(\lambda) = \det(A - \lambda I)$$ where $$\lambda$$ is the eigenvalue and $$I$$ is the identity matrix of the same size as $$A$$. First, compute $$A - \lambda I$$: $$A - \lambda I = \begin{pmatrix} 0 & 0 & 3 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{pmatrix} - \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} = \begin{pmatrix} -\lambda & 0 & 3 \\ 1 & -\lambda & 2 \\ 0 & 1 & 1 - \lambda \end{pmatrix}$$ Now calculate the determinant of this matrix: $$\det(A - \lambda I) = \det\begin{pmatrix} -\lambda & 0 & 3 \\ 1 & -\lambda & 2 \\ 0 & 1 & 1 - \lambda \end{pmatrix}$$ Using cofactor expansion along the first row: $$\det(A - \lambda I) = (-\lambda) \det\begin{pmatrix} -\lambda & 2 \\ 1 & 1 - \lambda \end{pmatrix} - 3 \det\begin{pmatrix} 1 & -\lambda \\ 0 & 1 \end{pmatrix}$$ The first determinant is: $$\det\begin{pmatrix} -\lambda & 2 \\ 1 & 1 - \lambda \end{pmatrix} = (-\lambda)(1 - \lambda) - (2)(1) = -\lambda + \lambda^2 - 2 = \lambda^2 - \lambda - 2$$ The second determinant is: $$\det\begin{pmatrix} 1 & -\lambda \\ 0 & 1 \end{pmatrix} = 1$$ Thus, the characteristic polynomial becomes: $$\det(A - \lambda I) = -\lambda (\lambda^2 - \lambda - 2) - 3(1) = -\lambda^3 + \lambda^2 + 2\lambda - 3$$ So the characteristic polynomial is: $$p_A(\lambda) = -\lambda^3 + \lambda^2 + 2\lambda - 3$$ ### Step 2: Find the minimal polynomial The minimal polynomial must divide the characteristic polynomial, and it is the polynomial of the smallest degree such that $$p(A) = 0$$, where $$A$$ is substituted into the polynomial. We can check factors of the characteristic polynomial to find the minimal polynomial. However, for a matrix of size $$3 \times 3$$, the minimal polynomial can be of degree 1, 2, or 3. Based on the structure of the matrix, the minimal polynomial is generally of degree 3, and in this case, it will match the characteristic polynomial: $$m_A(\lambda) = -\lambda^3 + \lambda^2 + 2\lambda - 3$$ Therefore, the minimal polynomial associated with the given matrix is: $$m_A(\lambda) = \lambda^3 - \lambda^2 - 2\lambda + 3$$ Would you like further details on this process or have any questions? Here are 5 related questions: 1. How is the characteristic polynomial related to the eigenvalues of a matrix? 2. What is the difference between the characteristic polynomial and the minimal polynomial? 3. How can we compute eigenvalues from the characteristic polynomial? 4. Why does the minimal polynomial have to divide the characteristic polynomial? 5. How do we verify if a given polynomial is the minimal polynomial? **Tip:** The minimal polynomial is always unique and gives important information about the diagonalizability of a matrix.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Linear Algebra
Minimal Polynomial
Characteristic Polynomial
Formulas
Characteristic Polynomial: p_A(\lambda) = \det(A - \lambda I)
Minimal Polynomial: m_A(\lambda)
Theorems
Cayley-Hamilton Theorem
Diagonalizability of Matrices
Suitable Grade Level
Undergraduate (Math major)
Related Recommendation
How to Calculate the Minimal Polynomial of an Element or Matrix
Cayley-Hamilton Theorem for 3x3 Matrix with Characteristic Polynomial
Diagonalization of a Matrix Using Orthogonal Transformation
Find Eigenvalues and Eigenvectors of a 3x3 Matrix - Step-by-Step Guide
Validating Determinant and Characteristic Polynomial Calculation