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# Help with this generalised eigenvector Watch

1. I was wondering if anyone could explain, from the following information that we put z=0, when I have always, when given the information that follows thought we could let z equal anything and so let z = 1.

We have, with:

So we can infer:

We also have:

So we still have the same equations as before.

I thought in a case like this that z could take any value so we let z = 1. But z = 0, can anyone explain why? I think I should point out that v_1 and v_2 need to be linearly independent, so if putting z = 0 makes them linearly independent, could someone explain why.

Thanks guys.
2. (Original post by TheBhramaBull)
.
Any value of z will give you the property you want for . However, letting z=0 makes orthogonal to , so it may be preferable to let z=0.

In reference to your linear independence bit at the end - do you know the definition of linear independence? If you can visualise how a set of linearly independent vectors in looks, then it should be fairly easy to see when and are linearly independent.

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Updated: December 24, 2010
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