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Quick Eigen Vector Question

  • Thread starter superkam
  • Start date
  • #1
17
0

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.
 

Answers and Replies

  • #2
vela
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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).
 
  • #3
17
0
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. :)
 

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