Eigen values and cubic roots question

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SUMMARY

The discussion centers on finding the roots of the characteristic equation λ³ - 2λ², which represents the eigenvalues of a matrix. The correct factorization is λ²(λ - 2), yielding roots of 0 (a double root) and 2. Participants emphasize that solving cubic equations can be challenging, suggesting methods such as factoring, guessing roots, or applying a cubic formula analogous to the quadratic formula, although the latter is less commonly known.

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  • Understanding of eigenvalues and characteristic equations
  • Familiarity with polynomial factorization techniques
  • Knowledge of the quadratic formula and its cubic counterpart
  • Basic algebraic manipulation skills
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  • Study polynomial factorization methods for cubic equations
  • Learn the cubic formula for finding roots of cubic equations
  • Practice solving characteristic equations for various matrices
  • Explore numerical methods for approximating roots of polynomials
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Students of linear algebra, mathematicians, and anyone involved in solving polynomial equations or studying matrix theory.

iamsmooth
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So I found the characteristic equation of a matrix, and I know the roots of the equation are supposed to be the eigenvalues. However, my equation is:

[tex]\lambda^3-2\lambda^2[/tex]

I have double checked different row expansions to make sure this answer is correct. So don't worry about how I came to get that equation.
I'm just not sure how to get roots from this. Would it be:

[tex]\lambda^2(\lambda-2)[/tex]

So that the roots are 0 and 2?

Basically I have trouble with cubic roots, I guess this is less of a question about eigenvalues than it is about cubic roots.

Thanks.
 
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You are right about your example (note 0 is a double root).
In general cubic equations are no fun. Your options are:
1. Factor, like you did above
2. Guess an answer [itex]\lambda_0[/itex], then divide out [itex](\lambda-\lambda_0)[/itex] and solve the resulting quadratic equation. This works well in constructed problems where you can easily see that a value like 0, 1, -1, 2 or -2 satisfies the equation.
3. Use the analog of the quadratic formula ([itex](-b\pm\sqrt{b^2-4ac})/2a[/itex]) for cubic equations. However it is messy, and unlike the quadratic formula hardy anyone knows it by heart.
 

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