A basis that makes a matrix triangular. Find a basis for $\mathbb C^3$ so that the following matrix is in triangular form:
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{pmatrix}
What are the eigenvalues?
I found the eigenvalues to be $\lambda_1 = 1, \lambda_2 = \frac{1}{2}(-1 - i \sqrt{3} ), \lambda_3 = \frac{1}{2}(-1 + i \sqrt{3} ).$ And the eigenvectors to be $$\begin{pmatrix}1\\
1\\
1
\end{pmatrix}, \begin{pmatrix}\frac{ -(\sqrt{3} + i)^2}{4}\\
\frac{ i(\sqrt{3} + i)}{2}\\
1
\end{pmatrix},\begin{pmatrix} \frac{ 2i}{\sqrt{3} - i}\\
\frac{ -2i}{\sqrt{3} + i}\\ 1
 \end{pmatrix}$$
But still I do not know how this will help me to find a basis for $\mathbb C^3$ so that the given matrix is in triangular form. I was told that I may find a dual basis, but still I do not know what exactly the steps I should do. Could anyone help me in solving this problem please? (not necessarily using dual basis and probably using an elegant and clean general way of doing this)

My calculations showed the following:
(Kindly try yourself and verify these results)
Eigenvalue: $\lambda_1 = 1$
Eigenvector for $\lambda_1 = 1$:
We solve the equation
$$
(A - \lambda_1 I) \mathbf{x} = \mathbf{0}
$$
The RREF of $A - \lambda_1 I$ is obtained as
$$
R_1 = \left[ \begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
\end{array} \right]
$$
Thus, we solve the equations
$$
x_1 - x_3 = 0, \ \ x_2 - x_3 = 0
$$
Easy to see that we get an eigenvector by taking $x_3$
as a free variable and setting $x_3 = 1$.
Thus, we get the eigenvector
$$\mathbf{v}_1 = \left[ \begin{array}{c}
1 \\
1 \\
1 \\
\end{array} \right] $$
Eigenvalue: $\lambda_2 = -{1 \over 2} - i {\sqrt{3} \over 2} $
Eigenvector for the eigenvalue $\lambda_2$:
We solve the equation
$$
(A - \lambda_2 I) \mathbf{x} = \mathbf{0}
$$
It is easy to verify that RREF of $A-\lambda_2 I$ is obtained as
$$
R_2 = \left[ \begin{array}{ccc}
1 & 0 & {1 \over 2} + i {\sqrt{3} \over 2} \\[2mm]
0 & 1 & {1 \over 2} - i {\sqrt{3} \over 2} \\[2mm]
0 & 0 & 0 \\[2mm]
\end{array} \right]
$$
Thus, we get an eigenvector as:
$$\mathbf{v}_2 = \left[ \begin{array}{c}
-{1 \over 2} - i {\sqrt{3} \over 2} \\
-{1 \over 2} + i {\sqrt{3} \over 2}  \\
1 \\
\end{array} \right] $$
Eigenvalue: $\lambda_3 = \bar{\lambda}_2 = -{1 \over 2} + i {\sqrt{3} \over 2} $
Eigenvector for $\lambda_3$:
We solve the equation
$$
(A - \lambda_3 I) \mathbf{x} = \mathbf{0}
$$
It is easy to see that RREF of $A - \lambda_3$ is obtained as
$$
R_3 = \left[ \begin{array}{ccc}
1 & 0 & {1 \over 2} - i {\sqrt{3} \over 2} \\[2mm]
0 & 1 & {1 \over 2} + i {\sqrt{3} \over 2} \\[2mm]
0 & 0 & 0 \\[2mm]
\end{array} \right]
$$
Hence, we get an eigenvector as:
$$\mathbf{v}_3 = \bar{\mathbf{v}}_2 = \left[ \begin{array}{c}
-{1 \over 2} + i {\sqrt{3} \over 2} \\
-{1 \over 2} - i {\sqrt{3} \over 2}  \\
1 \\
\end{array} \right] $$
Form the modal matrix:
$$
P = \left[ \matrix{ 
 \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \cr} \right]
$$
The eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are
linearly independent, since they correspond to different eigenvalues.
It is easy to verify that
$$
P^{-1} A P = D = \left[ \begin{array}{ccc}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3 \\
\end{array} \right]
$$
