Eigenvalues of a 4x4 matrix and the algebraic multipicities

[andrelutz001]

Hi everyone

Homework Statement

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. Homework Equations

det(A-λI)=0

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

 

[Dick]

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.

 

[andrelutz001]

Thank you