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anonymous
 5 years ago
let A=
(4 3 1
2 1 1
0 0 2)
a)find all eigenvalues of A.
b) for each of the eigenvalues of A find the corresponding eigenspace, and give a basis for each eigenspace. ... i mostly know how to do all of this just not sure on the basis part... any help ? im pretty sure the eigenvalues are 2,2,1?
anonymous
 5 years ago
let A= (4 3 1 2 1 1 0 0 2) a)find all eigenvalues of A. b) for each of the eigenvalues of A find the corresponding eigenspace, and give a basis for each eigenspace. ... i mostly know how to do all of this just not sure on the basis part... any help ? im pretty sure the eigenvalues are 2,2,1?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw A is a 3x3 matrix if you cant tell :S

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you familiar with paul's online notes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i totally agree with your comments on college

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as far as paying 40 k to teach yourself, lol. ironic huh

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got about 10 tabs opened up but very boring to read and im not very good at reading a bunch of boring stuff lol for that id be able to just read a textbook :S i have add...short attention span ftl

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0paul is boring? hes better than a text book in some sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and yep definitely but...makes for a nice degree lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think the future is teaching yourself, since we're doing it anyway

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as in, online supplements like khan academy and pauls notes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uhh yea idk i find it to be a bit bleh maybe more because i have... about 10 tabs of it opened that im trying to learn at once and skim through to find what i actually need to reteach myself lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0paul is a fluttering genius. but just with calculus and linear algebra. where are is his probability notes, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh cmon, he has all the examples done out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you want to do the eigenspace whachamakallit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we’re going to start with a square matrix A and try to determine vectors x and scalars so that we will have,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0scalar lambda such that A x = lamda * x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so in other words, multiplying the vector x by A is equivalent to multiplying the vector x by some scalar like 2 or 3 . remember x is a column vector, A is a square matrix, and they have to be defined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its like multiplying A by x is lengthening the arrow of x or dilating it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x is the eigenvector, and lambda the scalar is the eigenvalue

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok for the beginning im aware you subtract lambda I from the original matrix A then you take the determinant of that and set it = to 0 to find the eigenvalues of A

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you plug in each eigenvalue into the Alambda I but then what do you do for the basis after that? the way i was taught was marked wrong on the worksheet i have

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i had done is i REF the matrix after i plugged in the eigenvalue then i took thecolumn with leading nonzero terms... the only thing i can think of is maybe i had to take that column but from the original matrix?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we did A x = lambda I * x lambda I x  A x = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not necessary cuz you end up with 2 3 1 and 2 rows of 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well not for the eigenvalue of 2 i didnt try 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 you end up with 1  1 0 for the first row second row is 0 0 1 and then a row of 0s

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know how to find the basis of the kernal...basis of the range... but not sure on just the basis of the eigenspace lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thats ok a row of zeroes is fine brb, 10 minutes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kk im off to bed for the night ty anyway
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