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# yet another question concerning eigenvectors watch

1. you have the matrix

4 -3 0
1 3 2
-1 0 1

and are told that there is only one real eigenvalue (which can be found to be 3).

Then the question says to "find a normalised eigenvector corresponding to this eigenvalue"

How do I DOOOO this?

2. 4x - 3y = 3x . . . . . (1)
x + 3y + 2z = 3y . . . . . (2)
-x + z = 3z . . . . . (3)

(1) and (3) imply x = 3y and x = -2z.

(x, y, z) = k(6, 2, -3) satisifies (1), (2) and (3) for any k.

We choose k = 1/sqrt(6^2 + 2^2 + 3^2) = 1/7 so that k(6, 2, -3) is a vector of length 1.

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