Math  /  Algebra

Question Let A=[004040206]\text { Let } A=\left[\begin{array}{ccc} 0 & 0 & 4 \\ 0 & -4 & 0 \\ -2 & 0 & -6 \end{array}\right] (a) Determine the eigenvalues λ1\lambda_{1} and λ2\lambda_{2} of AA where λ1<λ2\lambda_{1}<\lambda_{2}. λ1=\lambda_{1}= \square λ2=\lambda_{2}= \square (b) Determine the algebraic multiplicity mm of each of the eigenvalues in part (a). m(λ1)=m\left(\lambda_{1}\right)= \square m(λ2)=m\left(\lambda_{2}\right)= \square

Studdy Solution
Determine the algebraic multiplicity of each eigenvalue:
The eigenvalues are λ1=4 \lambda_1 = -4 and λ2=2 \lambda_2 = -2 .
From the characteristic polynomial:
1. λ1=4 \lambda_1 = -4 appears twice (once from the linear factor and once from the quadratic factor).
2. λ2=2 \lambda_2 = -2 appears once.

Thus, the algebraic multiplicities are:
m(λ1)=2 m(\lambda_1) = 2 m(λ2)=1 m(\lambda_2) = 1
The eigenvalues and their multiplicities are: λ1=4,m(λ1)=2\lambda_1 = -4, \quad m(\lambda_1) = 2 λ2=2,m(λ2)=1\lambda_2 = -2, \quad m(\lambda_2) = 1

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