If the characteristic polynomial of a 3×3 matrix \(M\) over \(\mathbb{R}\)…
2017
If the characteristic polynomial of a 3×3 matrix \(M\) over \(\mathbb{R}\) (the set of real numbers) is \(\lambda^3 – 4 \lambda^2 + a \lambda +30, \quad a \in \mathbb{R}\), and one eigenvalue of \(M\) is 2, then the largest among the absolute values of the eigenvalues of \(M\) is _______
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Correct answer: 5
We are given the characteristic polynomial λ³ − 4λ² + aλ + 30 and told that one eigenvalue is 2. Use this to find a and factor the polynomial.
Since 2 is a root, plug λ = 2 into the polynomial and set it to zero: 8 − 16 + 2a + 30 = 0
Simplify to get 22 + 2a = 0, so a = −11.
With a = −11 the polynomial becomes λ³ − 4λ² − 11λ + 30. Divide by (λ − 2) to factor out the known root.
Polynomial division (or synthetic division) yields the quadratic quotient λ² − 2λ − 15.
Factor the quadratic: λ² − 2λ − 15 = (λ − 5)(λ + 3), so the eigenvalues are 2, 5, and −3.
Their absolute values are 2, 5, and 3, so the largest absolute value is 5.
Answer: 5