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Eigenvalues of a 4x4 matrix and the algebraic multipicities |
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| Sep16-09, 07:55 AM | #1 |
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Eigenvalues of a 4x4 matrix and the algebraic multipicities
Hi everyone
1. The problem statement, all variables and given/known data Consider the following 4 x 4 matrix: A = [[6,3,-8,-4],[0,10,6,7],[0,0,6,-3],[0,0,0,6]] Find the eigenvalues of the matrix and their multiplicities. Give your answer as a set of pairs: {[lambda1,multiplicity1],[lambda2,multiplicity2],........} 2. Relevant equations det(A-λI)=0 3. The attempt at a solution Set up the characteristics equation and solve it: A = [[6,3,-8,-4],[0,10,6,7],[0,0,6,-3],[0,0,0,6]] det(A-λI)= [[6-λ,3,-8,-4],[0,10-λ,6,7],[0,0,6-λ,-3],[0,0,0,6-λ]] This is the part where I think I am likely to make a mistake since it is rather difficult to factorize the characteristic polynomial using conventional methods(by hand). Therefore after a few steps the characteristic polynomial for the above matrix is: (λ^4)-(28*λ^3)+(288* λ^2)-(1296 *λ)+2160 Factorizing the characteristic polynomial yields: ((λ-10)(λ-6)^3) Looking at the problem statement again, the question asks to find the eigenvalues and the algebraic multiplicities. λ-10=0 therefore λ1=10 λ-6=0 therefore λ2=6 I know that the term algebraic multiplicity of an eigenvalue means the number of times it is repeated as a root of the characteristic equation. With this in mind I am inclined to state that for λ1=10 the algebraic multiplicity is 1 and for λ2=6 the algebraic multiplicity is 3. Therefore the answer as a set of pairs mentioned above {[lambda1,multiplicity1],[lambda2,multiplicity2],........} is {[10,1],[6,3]} I have made a good attempt at solving this question, am I on the right track? Thank you in advance, Andrei 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution |
| Sep16-09, 08:22 AM | #2 |
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I don't see any problems. But you did work too hard to find the characteristic polynomial. Your matrix is upper triangular. If you had used a determinant method like expansion by minors, you would have gotten the determinant to come out directly as (6-λ)*(10-λ)*(6-λ)*(6-λ). Only the diagonal elements contribute.
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| Sep16-09, 08:40 AM | #3 |
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Thank you
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