Question:
We have matrix $A$ defined as:
$$ A=\begin{bmatrix} -29 & 39 & -69 \\ -41 & 41 & -81 \\ -11 & 1 & -11 \end{bmatrix} $$
Eigenvectors for matrix $A$ are $$ v_1= \begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix} v_2=\begin{bmatrix}-1 \\ 1 \\ 1 \end{bmatrix} v_3=\begin{bmatrix}-3 \\ -2 \\ 1 \end{bmatrix} $$ If you would actually calculate eigenvalues and then eigenvectors for matrix $A$ you wouldn't get $v_1$, $v_2$ and $v_3$ as shown here. I guess the idea is to calculate eigenvalues if we were to make assumption that eigenvectors were these.
Attempt to solve:
First i've calculated characteristic polynomial for matrix A without solving eigenvalues from it. Characteristic polynomial is defined as:
$$ P_a(\lambda)=\det(A-\lambda I) $$ $$ P_a(\lambda)=\det\left(\begin{bmatrix}-29 & 39 & -69 \\ -41 & 41 & -81 \\ -11 & 1 & -11 \end{bmatrix}\times \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\lambda \right)$$ $$ P_a(\lambda)=\det \left(\begin{bmatrix}-29-\lambda & 39 & -69 \\ -41 & 41-\lambda & -81 \\ -11 & 1 & -11-\lambda \end{bmatrix}\right) $$ $$ P_a(\lambda)= -\lambda^3+\lambda^2+400\lambda -400 $$
Eigenvectors could be solved from equation. Calculating one vector per one $\lambda$
$$(A-\lambda I)v=0$$ $$Av-\lambda I v =0$$ $$\lambda I = Av$$ Now writing the equation in matrix form:
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix}=\begin{bmatrix} -29 & 39 & -69 \\ -41 & 41 & -81 \\ -11 & 1 & -11\end{bmatrix} \times v_1,v_2 \ldots v_3$$
But now how do you solve one $\lambda$ for one eigenvector $v_n$ where $n \in [1,2,3]$ and at this point I am starting to doubt if this even correct way to begin with ?
If someone could give hint towards correct solution to this problem that would be greatly appreciated.
Thanks,
Tuki