Find the characteristic roots and characteristic vectors of the matrix $\displaystyle A$ given by
$\displaystyle
A= \begin{bmatrix}
1 & 2 & 3 \\
0 & 2 & 3 \\
0 & 0 & 2
\end{bmatrix}
$
This matrix is already in diagonal form. You find the roots of the characteristic polynomial det(A-Ix). In this case, they are just the things on the diagonal.
To find the "characteristic vectors" I assume you are talking about the eigenvectors. Simply put the eigenvalue in for x in A-Ix (just subtract the value from the diagonal). Then you set that equal to 0 and find a basis for this solution space (the null space). This is the basis for the kernel of $\displaystyle (A-I\lambda)x$.
I have got $\displaystyle \lambda = 1,2$
Taking $\displaystyle \lambda = 1 $
$\displaystyle \begin{bmatrix} 0 & 2 & 3\\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}$ $\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ = $\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ $\displaystyle \Rightarrow$ $\displaystyle \begin{cases} 2y + 3z = 0 \\ y + 3z =0 \\z = 0 \end{cases}$
I am stuck here what should i do know from here onwards ?????
That should tell you that $\displaystyle y=0$, $\displaystyle z=0$ and $\displaystyle x$ is arbitrary. So take:
$\displaystyle \bold{x}_{(\lambda=1)}=\left[\begin{array}{c}1\\0\\0 \end{array}\right]$
as the eigenvector corresponding to eigen value $\displaystyle \lambda=1$
When you work this for the other eigen value (which has multiplicity 2) you will find two solutions for the eigen vector (defined up to an arbitrary constant).
CB
$\displaystyle \lambda = 2$
$\displaystyle \begin{bmatrix} -1 & 2 & 3 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix}$ $\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \begin{cases} -x + 2y + 3z = 0 \\ 3z = 0 \\ \end{cases}$
$\displaystyle \therefore$ $\displaystyle z = 0$
and $\displaystyle - x + 2y = 0$
U are correct so What should i do now ?????