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)
 A: 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
$$
P^{-1} A P = D = \left[ \begin{array}{ccc}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3 \\
\end{array} \right]
$$
