# Thread: Find the characteristic roots and characteristic vector

1. ## 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}
$

2. 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 $(A-I\lambda)x$.

3. 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 )

4. Whoa! whoops. Yeah exactly. it is 4:42 am here, I think it is bedtime, lol. Take care.

5. ## I am stuck here

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 )

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 ?????

6. 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

7. ## Is this what u are talking

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 ?????

8. 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
$\bold{x}_{(\lambda=2)}=\left[\begin{array}{c}2\\1\\0 \end{array}\right]$

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# find the characteristic roots and characteristic vectors of matrix

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