Quick Eigen Vector Question: Finding Eigen Vectors for Matrix A | Homework Help

[superkam]

Homework Statement

Hi, for the matrix A =
0 1 0

1 0 0

0 0 2

I have calculated the eigen values, and have successfully calculated the eigen vectors for lamda = -1 and 1. However for lamda = 2 the solutions says the answer is: Eigen vector: A( 0, 0, 1) where A is arbitrary and the vector is a column vector. The problem I am having is seeing where this answer came from.

Homework Equations

(A - lamda*I₃)*(X, Y, Z) (column vector) = 0

The Attempt at a Solution

As I said above I already know the correct Eigen vector for this particular Eigen value, I just cannot see where the answer has come from. From using the equation above I got the three following equations:
-2X + Y = 0
-2Y = 0
0 = 0

From these equations I cannot see where you get an eigen vector including a value for Z. Any help would be very much appreciated, thanks in advance, Kamran.

 

[vela]

Homework Statement

Hi, for the matrix A =
0 1 0

1 0 0

0 0 2

I have calculated the eigen values, and have successfully calculated the eigen vectors for lamda = -1 and 1. However for lamda = 2 the solutions says the answer is: Eigen vector: A( 0, 0, 1) where A is arbitrary and the vector is a column vector. The problem I am having is seeing where this answer came from.

Homework Equations

(A - lamda*I₃)*(X, Y, Z) (column vector) = 0

The Attempt at a Solution

As I said above I already know the correct Eigen vector for this particular Eigen value, I just cannot see where the answer has come from. From using the equation above I got the three following equations:
-2X + Y = 0
-2Y = 0
0 = 0

From these equations I cannot see where you get an eigen vector including a value for Z. Any help would be very much appreciated, thanks in advance, Kamran.

The second equation tells you y=0, and then the first equation tells you x=0. There's no condition on z, so z is arbitrary. In other words, the eigenvector is of the form (0,0,t)=t(0,0,1).

 

[superkam]

The second equation tells you y=0, and then the first equation tells you x=0. There's no condition on z, so z is arbitrary. In other words, the eigenvector is of the form (0,0,t)=t(0,0,1).

Ok I understand now, thank you for your help. :)