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Let \(A\) be a nonnegative irreducible \(3 \times 3\) matrix whose eigenvalues satisfy \(\lambda_{1}=2=\left|\lambda_{2}\right|=\left|\lambda_{3}\right|\) Determine \(\lambda_{2}\) and \(\lambda_{3}\)

Short Answer

Expert verified
In conclusion, for a nonnegative irreducible \(3 \times 3\) matrix A with eigenvalues \(\lambda_1 = 2\) and \(|\lambda_2| = |\lambda_3| = 2\), the eigenvalues are \(\lambda_2 = -2\) and \(\lambda_3 = -2\).

Step by step solution

01

Understand the properties of eigenvalues

Given that the matrix A is nonnegative and irreducible, it means that its eigenvalues will be real and nonnegative. Since 位鈧 = 2 and |位鈧倈 = |位鈧億 = 2, this implies that either 位鈧 and 位鈧 are positive or negative.
02

Use the Perron-Frobenius Theorem

According to the Perron-Frobenius Theorem for irreducible nonnegative matrices, the largest eigenvalue has a corresponding eigenvector with all positive entries. We already know that 位鈧 is the largest eigenvalue, so 位鈧 and 位鈧 must be smaller in absolute value.
03

Calculate the spectral radius

The spectral radius of a matrix is the maximum of the absolute values of its eigenvalues, and it can be denoted as 蟻(A). In our case, since 位鈧 is the largest eigenvalue and has an absolute value of 2, we have: \[蟻(A) = \max\{|位鈧亅, |位鈧倈, |位鈧億\} = 2\]
04

Analyze the possible values of 位鈧 and 位鈧

We know that 位鈧 and 位鈧 are real and have an absolute value of 2. This means that there are only two possibilities for each: 位鈧 = 2 or 位鈧 = -2, and 位鈧 = 2 or 位鈧 = -2. However, 位鈧 = 2, so 位鈧 and 位鈧 must be distinct from 位鈧. Therefore, we conclude that 位鈧 = -2 and 位鈧 = -2. In conclusion, the eigenvalues 位鈧 and 位鈧 of the nonnegative irreducible matrix A are both -2.

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