Technical problem with eigenvalues

[Yankel]

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:

$$-\lambda ^{3}+8\lambda ^{2}-20\lambda +16$$

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:

$$-\lambda ^{3}+8\lambda ^{2}-20\lambda +16 = 0$$

I mean, how do I expand this polynomial into:

$$(\lambda -4)(\lambda -2)^{2}$$

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:

$$\begin{pmatrix} 3 &2 &3 \\ -1 &0 &-3 \\ 1 &2 &5 \end{pmatrix}$$thanks !

 

[Ackbach]

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]

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:

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