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# Linear Algebra Foundations #10 - Eigenvectors

# Linear Algebra Foundations #10 - Eigenvectors

bennattj + 0 comments Well this "notation" couldn't have been more confusing. For anyone else confused, this is what they are asking for:

The k_1 and k_2 (or both k_1's as they wrote it) are simply the fact that a multiple of an eigenvector is also an eigenvector.

bookidodo + 0 comments There is a mistake in the data. k1 =\= "the other" k1

Someone please change this to k1 and k2 as suggested above.

The method that is to be taken if k1 and the other "k1" were equal is going to find the eigenspace of this k1. But, alas, when these are not the same one: that's a FATAL mistake. (Sorry).

merajmasuk + 1 comment What is actually k1 here?

octavianarsene + 0 comments k is the positive integer variable (consider it as a multiple)

Check this: https://www.safaribooksonline.com/library/view/linear-algebra-concepts/9781107484535/16_chapter-title-8.html

shubhamkr11 + 0 comments I could not understand the formula provided - v1 = k1 [+1 A ]T

kiner_shah + 1 comment Ans is -2 -1 right? I am getting WA!

sagarys49Asked to answer + 2 comments **Endorsed by kiner_shah**yes it is correct but you are not putting in the correct form type each solution in the seperate line.

you will ge tthe answer correctly.

kiner_shah + 0 comments Got it! Thnxx! :-)

sahmad39 + 1 comment Isn't there 2 eigenvectors? One for when lambda = -1 and one for lambda = -2? So like one is (-1,1) and the other is (-2,-1) ?

ryklin + 1 comment I got span[-1 1] for lambda -1 and span [-2 1] for lambda -2

I think in the lambda -2 case you might have made a mistake because row reduced echelon form of the matrix is: [1 1/2; 0 0]

Therefore if v1 == - 1/2*v2, you then multiply by -2 and you get [-2 1] as the solution for the vector v.

I double checked this solution.

Nonetheless, I have no idea how to enter this into the solution box!

thebick + 0 comments see what bennattj wrote: A and B fill the column vectors (1 A) and (1 B).

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