Calculate eigenvalues and eigenvectors I have this matrix below, and I need to find the eigenvalues.\begin{bmatrix}
        2 & -1 & 0 \\
        5 & 2 & 4 \\
        0 & 1 & 2 \\
        \end{bmatrix}
This is what I have done so far:
I used $\det(A-λI)=0$ and reached this form \begin{bmatrix}
        2-λ & -1 & 0 \\
        5 & 2-λ & 4 \\
        0 & 1 & 2-λ \\
       \end{bmatrix}
I have done some simplifications:
$(2-λ)[(2-λ)(2-λ)-4]-5(-(2-λ))=0$
$(2-λ)[(2-λ)(2-λ)-4]-5(-1)=0$
$(2-λ)[4-4λ+λ^2-4+5]=0$
$(2-λ)[λ^2-4λ+5]=0$
$λ^2(2-λ)-4λ(2-λ)+5(2-λ)=0$
$2λ^2-λ^3-8λ+4λ^2+10-5λ=0$
$-λ^3+6λ^2-13λ+10=0$
or $-λ(λ^2-6λ+13)+10=0$
$-λ (λ-(3+{\sqrt 22})) (λ+(3-{\sqrt 22}))+10=0$
Am I doing it right and if so:
I checked the answer on Symbolab and it was $2,2-i,2+i$, How so? and what 
is $i$? 
And is the matrix will be like this when I want to calculate the eigenvector for $2-i$ ??\begin{matrix}
        i & -1 & 0 \\
        5 & i & 4 \\
        0 & 1 & i \\
       \end{matrix}

It happens that, if $\lambda=2$, then $-\lambda^3+6\lambda^2-13\lambda+10=0$. In fact, $-\lambda^3+6\lambda^2-13\lambda+10=(2-\lambda)(\lambda^2-4\lambda+5)$. The roots of the polynomial $\lambda^2-4\lambda+5$ are complex non-real numbers: $2+i$ and $2-i$.
