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}

$

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- May 3rd 2009, 10:27 PMzorroFind the characteristic roots and characteristic vector
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}

$ - May 4th 2009, 01:10 AMGamma
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 $\displaystyle (A-I\lambda)x$. - May 4th 2009, 01:28 AMmr fantastic
- May 4th 2009, 01:43 AMGamma
Whoa! whoops. Yeah exactly. it is 4:42 am here, I think it is bedtime, lol. Take care.

- Dec 19th 2009, 10:23 PMzorroI am stuck here

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 ????? - Dec 19th 2009, 10:42 PMCaptainBlack
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 - Dec 20th 2009, 12:39 AMzorroIs this what u are talking

$\displaystyle \lambda = 2$

$\displaystyle \begin{bmatrix} -1 & 2 & 3 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix}$ $\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \begin{cases} -x + 2y + 3z = 0 \\ 3z = 0 \\ \end{cases}$

$\displaystyle \therefore$ $\displaystyle z = 0$

and $\displaystyle - x + 2y = 0$

U are correct so What should i do now ????? - Dec 20th 2009, 01:49 AMdedust
let $\displaystyle y=t$, $\displaystyle t \in \mathbb{R}$, then $\displaystyle x=2t$, and the solution is $\displaystyle (x,y,z) = (2t, t, 0) = t(2, 1, 0)$

so, the eigen vector is

$\displaystyle \bold{x}_{(\lambda=2)}=\left[\begin{array}{c}2\\1\\0 \end{array}\right]$