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Technical problem with eigenvalues

Yankel

Active member
Jan 27, 2012
398
Hello

I was trying to find eigenvalues of a matrix. I calculated the characteristic polynomial by calculating (A-lambdaI) and then calculating it's determinant. The results was:

[tex]-\lambda ^{3}+8\lambda ^{2}-20\lambda +16[/tex]

which is the correct calculation.

Now, the eigenvalues are 2,2,4, but I do not know, technically, how am I suppose to find it from:

[tex]-\lambda ^{3}+8\lambda ^{2}-20\lambda +16 = 0[/tex]

I mean, how do I expand this polynomial into:

[tex](\lambda -4)(\lambda -2)^{2}[/tex]

assuming that I do not see it in my eyes immediately, is there some hint to look for ?

Just to supply all information, the matrix is:

[tex]\begin{pmatrix} 3 &2 &3 \\ -1 &0 &-3 \\ 1 &2 &5 \end{pmatrix}[/tex]


thanks !!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
The first place to look would be the Rational Root Theorem. Then, once you get one root, you can perform polynomial long division or synthetic division to obtain a quadratic times the linear factor you just discovered.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Sometimes (this is the case) we can get factorized the characteristc polynomial with adequate transformations. For example, $R_3+R_2$ and $C_2-C_3$ provide:

$$\begin{aligned}\begin{vmatrix}{3-\lambda}&{2}&{3}\\{-1}&{-\lambda}&{-3}\\{1}&{2}&{5-\lambda}\end{vmatrix}&=\begin{vmatrix}{3-\lambda}&{2}&{3}\\{-1}&{-\lambda}&{-3}\\{0}&{2-\lambda}&{2-\lambda}\end{vmatrix}\\ &=\begin{vmatrix}{3-\lambda}&{-1}&{3}\\{-1}&{-\lambda+3}&{-3}\\{0}&{0}&{2-\lambda}\end{vmatrix}\\&=(2-\lambda)(\lambda^2-6\lambda+8)\\&=-(\lambda-2)^2(\lambda-4)\end{aligned}$$