Find the characteristic roots and characteristic vector

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May 3rd 2009, 10:27 PM
zorro

Find the characteristic roots and characteristic vector

Find the characteristic roots and characteristic vectors of the matrix $A$ given by

$ A= \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 0 & 2 \end{bmatrix} $
May 4th 2009, 01:10 AM
Gamma

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 $(A-I\lambda)x$ .
May 4th 2009, 01:28 AM
mr fantastic

Quote:

Originally Posted by Gamma

This matrix is already in diagonal form. [snip]

I'm sure you meant to say upper triangular form.

(By the way, welcome and thanks for all your help so far at MHF (Clapping))
May 4th 2009, 01:43 AM
Gamma

Whoa! whoops. Yeah exactly. it is 4:42 am here, I think it is bedtime, lol. Take care.
December 19th 2009, 10:23 PM
zorro

I am stuck here

Quote:

Originally Posted by mr fantastic

I'm sure you meant to say upper triangular form.

(By the way, welcome and thanks for all your help so far at MHF (Clapping))

I have got $\lambda = 1,2$

Taking $\lambda = 1$

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

I am stuck here what should i do know from here onwards ?????
December 19th 2009, 10:42 PM
CaptainBlack

Quote:

Originally Posted by zorro

I have got $\lambda = 1,2$

Taking $\lambda = 1$

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

I am stuck here what should i do know from here onwards ?????

That should tell you that $y=0$ , $z=0$ and $x$ is arbitrary. So take:

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

as the eigenvector corresponding to eigen value $\lambda=1$

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
December 20th 2009, 12:39 AM
zorro

Is this what u are talking

Quote:

Originally Posted by CaptainBlack

That should tell you that $y=0$ , $z=0$ and $x$ is arbitrary. So take:

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

as the eigenvector corresponding to eigen value $\lambda=1$

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

$\lambda = 2$

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

$\therefore$ $z = 0$

and $- x + 2y = 0$

U are correct so What should i do now ?????
December 20th 2009, 01:49 AM
dedust

Quote:

Originally Posted by zorro

$\lambda = 2$

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

$\therefore$ $z = 0$

and $- x + 2y = 0$

U are correct so What should i do now ?????

let $y=t$ , $t \in \mathbb{R}$ , then $x=2t$ , and the solution is $(x,y,z) = (2t, t, 0) = t(2, 1, 0)$

so, the eigen vector is
$\bold{x}_{(\lambda=2)}=\left[\begin{array}{c}2\\1\\0 \end{array}\right]$