# Thread: are these eigenvectors right?

1. ## are these eigenvectors right?

Note... I am fairly certain that my characteristic equation is correct. I just need someone to look at the eigenvectors to tell me if I have them correct. I included the matrix A, the lambdas and the rref for background. Thanks. I had an original 3x3 matrix A of:

[1 2 -1]
[2 1 1]
[-1 1 0]

for lambda 3 i get
[-2 2 -1]
[2 -2 1 ]
[-1 1 -3] rref to:

[1 -1 0]
[0 0 1]
[0 0 0]

so x - y = 0
z = 0
vector is
[1]
[1]
[0]

lamda = -2
[-3 2 -1 ]
[2 -3 1]
[-1 1 2]

rref
[1 0 0]
[0 1 0]
[0 0 1]

i get for the vector
[0]
[0]
[0]

lambda = 1
[0 2 -1]
[2 0 1]
[-1 1 -1]

rref
[1 0 1/2]
[0 1 -1/2]
[0 0 0]

let z = t
R1: x = -1/2t
R2: y = 1/2t
pick t = 2
[-1]
[1]
[2]

2. I put your matrix into MATLAB and got the following matrix of eigen vectors (where the columns are the eigenvectors)

-0.5774 -0.4082 -0.7071
0.5774 0.4082 -0.7071
-0.5774 0.8165 0.0000

Hope this helps

3. You can check them yourself just by multiplying.

$\begin{bmatrix}1 & 2 & -1 \\ 2 & 1 & 1 \\1 & 1 & 0 \end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}3 \\ 3 \\ 2\end{bmatrix}$
Since that is NOT 3 times $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ (which would be $\begin{bmatrix} 3 \\ 3 \\ 0\end{bmatrix}$) it is NOT an eigenvector corresponding to eigenvalue 3.

( $\begin{bmatrix}-2 & 2 & -1 \\ 2 & -2 & 1 \\ -1 & -1 & -3\end{bmatrix}$ row reduces to $\begin{bmatrix}1 & 0 & \frac{-5}{4} \\ 0 & 1 & \frac{-7}{4} \\ 0 & 0 & 0\end{bmatrix}$ which gives x- 5/4 z= 0, y- 7/4 z= 0 If you take z= 4, then x= 5 and y= 7 so that $\begin{bmatrix}5 \\ 7 \\ 4\end{bmatrix}$ is an eigenvalue. But I prefer to work directly from the definition of "eigenvalue" and "eigenvector".

If $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ is an eigenvector of A corresponding to eigenvalue 3, we must have $\begin{bmatrix}1 & 2 & -1 \\ 2 & 1 & 1 \\1 & 1 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\z\end{bmatrix}= \begin{bmatrix}3x \\ 3y \\ 3z\end{bmatrix}$ which gives the three equations x+ 2y- z= 3x, 2x+ y+ z= 3y, x+ y= 3z or -2x+ 2y- z= 0, 2x- 2y+ z= 0, and x+ y- 3z= 0, which corresponds to the matrix you give. The first two equations are the same. If we multiply the last equation by two and add to the second equation, y is eleminated: 4x- 5z= 0 so x= (5/4)z as we got before. )

$\begin{bmatrix}1 & 2 & -1 \\ 2 & 1 & 1 \\-1 & 1 & 0 \end{bmatrix}\begin{bmatrix}-1 \\ 1 \\ 2\end{bmatrix}= \begin{bmatrix}-1 \\ 1 \\ 2\end{bmatrix}$ which is 1 times the vector so this is an eigenmvector corresponding to eigenvalue 1.

You should know that the definition of "eigenvector corresponding to eigenvalue $\lambda$ is a non-zero vector, v, such that $Av= \lambda v$". The 0 vector, $\begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$ is NEVER an eigenvector.

The equations for eigenvector -2 should be x+ 2y- z= -2x, 2x+ y+ z= -2y, and x+ y= -2z or 3x+ 2y- z= 0, 2x+ 3y+ z= 0, and x+ y+ 2z= 0. Adding the first two equations gives 5x+ 5y= 0 or y= -x. Putting y= -x in the second equation, 2x- 3x+ z= -x+ z= 0 so z= x. Putting y= -x, z= x into the third equation, we get x- x+ 2x= 2x= 0 so that x= y= 0 is the only vector satifying that: there is NO non-zero vector, v, satisfying Av= -2v so -2 is NOT an eigenvalue! You had better recalculate that! Since the other two eigenvalues are correct, your characteristic equation is probably correct and knowing two eigenvalues it should be easy to find the third- recheck it.