Eigenvalue by Gaussian elimination?

Question

Let $A=\begin{bmatrix}-1&4&0\\-2&4&-1\\2&-5&0 \end{bmatrix}$

I want to find out the eigenvalue of matrix my Gaussian elimination. My idea was to make the matrix triangular and by that being able to write an expression for the determinant that equals 0. Like this: $(aλ+a_1)(bλ+b_1)(cλ+c_1)$

My try:

$ \det \begin{vmatrix} λ+1&-4&0\\2&λ-4&1\\-2&5&λ \end{vmatrix} = \begin{vmatrix} λ+1&0&0\\2&λ-4+\frac{8}{(λ+1)}&1\\-2&5-\frac{8}{(λ+1)}&λ \end{vmatrix}= \begin{vmatrix} λ+1&0&0\\2&λ-4+\frac{8}{(λ+1)}&0\\-2&5-\frac{8}{(λ+1)}&\frac{5-\frac{8}{(λ+1)}}{-4+\frac{8}{(λ+1)}} \end{vmatrix}$,

which would give the equation: $ (λ+1)(λ-4+\frac{8}{(λ+1)})(\frac{5-\frac{8}{(λ+1)}}{-4+\frac{8}{(λ+1)}})=0 $

Now solving each of the parentheses for λ, should give me the right Eigenvalues if I understand the concept of eigenvalues. Am I doing anything wrong or do i simply not understand eigenvalues yet?

Answer 1

Here is a simple way to obtain the characteristic polynomial: \begin{align} &\begin{vmatrix} \lambda+1&-4&0\\ 2&\lambda- 4&\phantom{-}\\-2&5&\lambda \end{vmatrix} = \begin{vmatrix} \lambda+1&-4&0\\ 0&\lambda+1&\lambda+1\\-2&5&\lambda \end{vmatrix} = (\lambda+1)\begin{vmatrix} \lambda+1&-4&0\\ 0 & 1 & 1\\-2&5&\lambda \end{vmatrix}\\[1ex] ={} (\lambda+1)&\begin{vmatrix} \lambda+1&-4&4\\ 0 & 1 & 0\\-2&5&\lambda -5 \end{vmatrix} = (\lambda+1)\begin{vmatrix} \lambda+1&4 \\-2&\lambda -5 \end{vmatrix}=(\lambda+1)\bigl((\lambda+1)(\lambda-5)+8\bigr)\\[1ex] =(\lambda+1)&(\lambda^2-4\lambda+3)=(\lambda+1)(\lambda-1)(\lambda-3). \end{align}