# Technical problem with eigenvalues

#### Yankel

##### Active member
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

##### Indicium Physicus
Staff member
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
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}