Question

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?

Answer 1

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

Answer 2

hey

Answer 3

hello

Answer 4

you familiar with paul's online notes

Answer 5

yep

Answer 6

nice

Answer 7

i totally agree with your comments on college

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

as in, online supplements like khan academy and pauls notes

Answer 14

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

Answer 15

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

Answer 16

oh cmon, he has all the examples done out

Answer 17

you want to do the eigenspace whachamakallit?

Answer 18

lol yes

Answer 19

alright, one sec

Answer 20

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

Answer 21

scalar lambda such that A x = lamda * x

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

yes

Answer 27

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

Answer 28

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?

Answer 29

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

Answer 30

why not RREF ?

Answer 31

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

Answer 32

also not possible

Answer 33

well not for the eigenvalue of 2 i didnt try 1

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

kk

Answer 38

kk im off to bed for the night ty anyway

Answer 39

ok