Problem
Find $3\times3$ matrix $\textbf{A}$ that has eigenvalues of $\lambda_1=-1,\lambda_2=1,\lambda_3=2$ and corresponding eignevectors $x_1=\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix},x_2=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix},x_3=\begin{bmatrix}1 \\2 \\1\end{bmatrix}$
Attempt to solve
I try to compute generic $3\times 3$ matrix and solve it's eigenvalues. We can compute eigenvalues via characteristic polynomial which is defined as:
$$ P_A(\lambda) := \det(\textbf{A}-\lambda\textbf{I}) $$
$$ P_A(\lambda)=\det(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}-\begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}) $$
$$=\begin{vmatrix} a-\lambda & b & c \\ d & e-\lambda & f \\ g & h & i-\lambda \end{vmatrix}$$
$$=(a-\lambda)\begin{vmatrix} e-\lambda & f \\ h & i-\lambda \end{vmatrix}-b\begin{vmatrix} d & f \\ g & i-\lambda \end{vmatrix} + c \begin{vmatrix} d & e-\lambda \\ g & h \end{vmatrix} $$
$$ =(a-\lambda)[(e-\lambda)(i-\lambda)-fh)]-b[d(i-\lambda)-fg]+c[dh-g(e-\lambda)] $$
$$ =(a-\lambda)[ei+e(-\lambda)-\lambda i -\lambda (-\lambda)-fh]-b[di-d\lambda-fg]+c[dh-ge-g\lambda] $$
$$= (a-\lambda)[\lambda^2-i\lambda-e\lambda+ei]-b[-d\lambda-di-fg]+c[-g\lambda-+dh-ge] $$
$$ = a\lambda^2-ia\lambda-ea\lambda+eai-\lambda^3+i\lambda^2+e\lambda^2-ei\lambda +bd\lambda + bdi + bfg +cdh -cge-cg\lambda $$
$$ P_A(\lambda) = -\lambda^3+\lambda^2(a+i+e)+\lambda(-ia-ea-ei+bd-cg)+eai+bdi+bfg+cdh-cge $$
Eigenvalues can be found when $P_A(\lambda)=0$. Only problem is I'am not quite sure if this is the best approach for this problem.
Any suggestion on how to proceed / change the approach all together.
Best Answer
$A=\begin{pmatrix}0&1&1\\1&1&2\\2&0&1\end{pmatrix}\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}\begin{pmatrix}0&1&1\\1&1&2\\2&0&1\end{pmatrix}^{-1}$.