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Chapter 6: Problem 9

Let \(A\) be a \(4 \times 4\) matrix and let \(\lambda\) be an eigenvalue of multiplicity 3. If \(A-\lambda I\) has rank 1 , is \(A\) defective? Explain.

Short Answer

Expert verified
Matrix A is not defective because the dimension of the eigenspace associated with the eigenvalue 位 is 3, which is equal to its algebraic multiplicity. This means that A has the same number of linearly independent eigenvectors as its algebraic multiplicity.

Step by step solution

01

Understanding Defective Matrices

A matrix is said to be defective if it has fewer linearly independent eigenvectors than its algebraic multiplicity. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic equation of the matrix. In this case, the algebraic multiplicity of 位 is 3, since it is given as an eigenvalue of multiplicity 3.
02

Determine the Nullspace of (A - 位I)

To find the eigenspace associated with the eigenvalue 位, we need to find the nullspace of the matrix (A - 位I). In this problem, we are specifically given that the rank of (A - 位I) is 1. Since the matrix is a 4x4 matrix, we can use the Rank-Nullity theorem: Rank(A - 位I) + Nullity(A - 位I) = 4 We are given Rank(A - 位I) = 1, therefore we have: 1 + Nullity(A - 位I) = 4 Nullity(A - 位I) = 3
03

Checking if A is Defective

Now, we have found the nullity of (A - 位I), which is 3. Since the nullity represents the dimension of the eigenspace associated with eigenvalue 位, we can conclude that the dimension of this eigenspace is 3. Since the algebraic multiplicity of 位 is also 3, which is equal to the dimension of its eigenspace, we have the same number of linearly independent eigenvectors as its algebraic multiplicity. Therefore, we can conclude that matrix A is not defective.

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