Thread: Find eigenvalues and eigenvector

I am to find the eigenvalues and corresponding eigenvectors of the following matrix:

[3 -1 -1]
[-12 0 5]
[4 -2 -1]

Here is what I have so far:
det(I - A) = det [-3 1 1], [12 -5], [-4 2 +1] = (-3)()(+1)


I know that this is not completely correct, but not sure how to finish. Thanks in advance.

your determinant does not look correct. i get you should have:

(λ-3)(λ)(λ+1) + (1)(-5)(-4) + (1)(12)(2) - (-4)(λ)(1) - (-5)(2)(λ-3) - (λ+1)(1)(12)

= λ^3 - 2λ^2 - 3λ + 20 + 24 + 4λ + 10λ - 30 - 12λ - 12

= λ^3 - 2λ^2 - λ + 2

Originally Posted by page929

I am to find the eigenvalues and corresponding eigenvectors of the following matrix:

[3 -1 -1]
[-12 0 5]
[4 -2 -1]

Here is what I have so far:
det(I - A) = det [-3 1 1], [12 -5], [-4 2 +1] = (-3)()(+1)


I know that this is not completely correct, but not sure how to finish. Thanks in advance.

$\displaystyle A=\begin{pmatrix} 3& -1 & -1\\ -12& 0 & 5\\ 4& -2 & -1\end{pmatrix}$

First we will find the characteristic polynomial- $\displaystyle \Delta (t)$

$\displaystyle \Delta (t)=det(tI-A)=det\begin{pmatrix} t-3& -1 & -1\\ -12& t & 5\\ 4& -2 & t+1\end{pmatrix}=t^3-2t^2-t+2=(t-2)(t-1)(t+1)$

Thus, our eigenvalues are:

$\displaystyle \lambda_1=2$

$\displaystyle \lambda_2=1$

$\displaystyle \lambda_3=-1$


Now, I'll find eigenvector for eigenvalue - $\displaystyle \lambda_1=2$ (And you will do the rest...)

$\displaystyle M:= \begin{pmatrix} 3& -1 & -1\\ -12& 0 & 5\\ 4& -2 & -1\end{pmatrix}-I\lambda_1= \begin{pmatrix} 3& -1 & -1\\ -12& 0 & 5\\ 4& -2 & -1\end{pmatrix}-2I=\begin{pmatrix} 3-2& -1 & -1\\ -12& 0-2 & 5\\ 4& -2 & -1-2\end{pmatrix}=\begin{pmatrix}1& -1 & -1\\ -12& -2 & 5\\ 4& -2 & -3\end{pmatrix}$

The eigenvectors that belongs to eigenvalue $\displaystyle \lambda_1=2$ creating the solution for homogeneous system $\displaystyle MX=0$, in other words:

$\displaystyle \begin{pmatrix}1& -1 & -1\\ -12& -2 & 5\\ 4& -2 & -3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0$

Now, you just need to solve it...

The solution $\displaystyle (x_1,y_1,z_1)$ is your eigenvector that belongs to eigenvalue $\displaystyle \lambda_1=2$ .

λ=2: (λI-A)=0
[(-1,1,1), (12,2,-5), (-4,2,3)] [x1, x2, x3] = 0
-x1 + x2 + x3 = 0
x1 = 2
x2 = 1
x3 = 1

λ=1: (λI-A)=0
[(-2, 1, 1), (12, 1, -5), (-4, 2, 2)] [x1, x2, x3] = 0
-2x1 + x2 + x3 = 0
x1 = 1
x2 = 1
x3 = 1

λ=-1: (λI-A)=0
[(-4, 1, 1), (12, -1, -5), (-4, 2, 0)] [x1. x2. x3] = 0
-4x1 + x2 + x3 = 0
x1 = 1
x2 = 2
x3 = 2

Are these eigenvector correct?

I alos need help on finding the eigenvector and want some clarity.

Are you suppose to just multiply the 3x3 and the 3x1 matrix to get a 3x1 matrix or are you suppose to take the parametric vector form?

Also shouldn't the solution be (x1, y1, z1)^T?

ANd what is x1, y1, and z1? Are they our respective eigenvalues?

And what do you do when there is a repeated eigenvalue? Does that mean there will be two of the same eigenvectors?