Math Problem Statement

Show that the matrix has less than three lineraly independent eigenvalues 3 10 5 -2 -3 -4 3 5 7

Solution

Let's consider the matrix AA given by:

3 & 10 & 5 \\ -2 & -3 & -4 \\ 3 & 5 & 7 \end{pmatrix}$$ To determine how many linearly independent eigenvalues the matrix has, we need to calculate its **characteristic polynomial** and find its roots. The characteristic polynomial $$p(\lambda)$$ is defined as: $$p(\lambda) = \det(A - \lambda I)$$ Where $$I$$ is the identity matrix and $$\lambda$$ is an eigenvalue. Let's compute this step-by-step. ### Step 1: Subtract $$\lambda I$$ from matrix $$A$$ We subtract $$\lambda$$ from the diagonal elements of $$A$$: $$A - \lambda I = \begin{pmatrix} 3 - \lambda & 10 & 5 \\ -2 & -3 - \lambda & -4 \\ 3 & 5 & 7 - \lambda \end{pmatrix}$$ ### Step 2: Calculate the determinant $$\det(A - \lambda I)$$ We now compute the determinant of the matrix $$A - \lambda I$$. $$\det(A - \lambda I) = \det\begin{pmatrix} 3 - \lambda & 10 & 5 \\ -2 & -3 - \lambda & -4 \\ 3 & 5 & 7 - \lambda \end{pmatrix}$$ Let's compute this determinant. The determinant of a 3x3 matrix is given by: $$\det(A - \lambda I) = (a_{11})\det\begin{pmatrix}a_{22} & a_{23}\\a_{32} & a_{33}\end{pmatrix} - (a_{12})\det\begin{pmatrix}a_{21} & a_{23}\\a_{31} & a_{33}\end{pmatrix} + (a_{13})\det\begin{pmatrix}a_{21} & a_{22}\\a_{31} & a_{32}\end{pmatrix}$$ ### Step 3: Solve the characteristic equation The roots of the characteristic polynomial give us the eigenvalues. We want to see how many distinct eigenvalues we have (i.e., how many linearly independent eigenvalues). Let me compute this determinant to finish the proof.The characteristic polynomial of the matrix $$A$$ is: $$p(\lambda) = -\lambda^3 + 7\lambda^2 - 16\lambda + 12$$ ### Step 4: Find the eigenvalues To determine the eigenvalues, we need to find the roots of this characteristic polynomial: $$-\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0$$ Let me solve this cubic equation to find the eigenvalues.The eigenvalues of the matrix $$A$$ are $$\lambda = 2$$ and $$\lambda = 3$$. Since we only have **two distinct eigenvalues**, and fewer than three linearly independent eigenvalues, this matrix does indeed have less than three linearly independent eigenvalues. Would you like a more detailed explanation or have any questions? Here are five related questions you might find helpful: 1. How do we interpret the geometric multiplicity of an eigenvalue? 2. What is the significance of repeated eigenvalues? 3. How can we determine if a matrix is diagonalizable based on its eigenvalues? 4. How do eigenvalues relate to the rank of a matrix? 5. How does the characteristic polynomial help in determining the matrix's invertibility? **Tip:** A matrix with fewer linearly independent eigenvalues than its size (here 3x3) may not be diagonalizable, and this can indicate special geometric properties.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues
Characteristic Polynomial

Formulas

Characteristic polynomial: p(λ) = det(A - λI)
Determinant of a 3x3 matrix
Cubic equation solving

Theorems

Eigenvalue Theorem
Diagonalizability Theorem

Suitable Grade Level

Undergraduate Level (Linear Algebra)