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

- chestercat

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- anonymous

btw A is a 3x3 matrix if you cant tell :S

- anonymous

hey

- anonymous

hello

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## More answers

- anonymous

you familiar with paul's online notes

- anonymous

yep

- anonymous

nice

- anonymous

i totally agree with your comments on college

- anonymous

as far as paying 40 k to teach yourself, lol. ironic huh

- anonymous

got 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

paul is boring? hes better than a text book in some sense

- anonymous

and yep definitely but...makes for a nice degree lol

- anonymous

i think the future is teaching yourself, since we're doing it anyway

- anonymous

as in, online supplements like khan academy and pauls notes

- anonymous

uhh 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 re-teach myself lol

- anonymous

paul is a fluttering genius. but just with calculus and linear algebra. where are is his probability notes, lol

- anonymous

oh cmon, he has all the examples done out

- anonymous

you want to do the eigenspace whachamakallit?

- anonymous

lol yes

- anonymous

alright, one sec

- anonymous

we’re going to start with a square matrix A and try to determine vectors x and scalars so that we will have,

- anonymous

scalar lambda such that A x = lamda * x

- anonymous

so 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

so its like multiplying A by x is lengthening the arrow of x or dilating it

- anonymous

x is the eigenvector, and lambda the scalar is the eigenvalue

- anonymous

ok 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

yes

- anonymous

then you plug in each eigenvalue into the A-lambda 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

what i had done is i REF the matrix after i plugged in the eigenvalue then i took thecolumn with leading non-zero terms... the only thing i can think of is maybe i had to take that column but from the original matrix?

- anonymous

we did A x = lambda I * x
lambda I x - A x = 0

- anonymous

why not RREF ?

- anonymous

not necessary cuz you end up with 2 -3 1 and 2 rows of 0

- anonymous

also not possible

- anonymous

well not for the eigenvalue of 2 i didnt try 1

- anonymous

1 you end up with 1 - 1 0 for the first row second row is 0 0 1 and then a row of 0s

- anonymous

i 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

ok thats ok
a row of zeroes is fine
brb, 10 minutes

- anonymous

kk

- anonymous

kk im off to bed for the night ty anyway

- anonymous

ok

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