Question

does the following set form a vector space?

Answer 1

http://assets.openstudy.com/updates/attachments/54054c46e4b0f2ed1e14b2a3-best_mathematician-1409633357762-untitled.png

Answer 2

yes it does since they are linearly independant

Answer 3

which one?

Answer 4

a) is a vector space

Answer 5

a set of linear independent vectors does not form a vector space always

Answer 6

yep, I do agrre and also second one is not but how?

Answer 7

i.e. {<1,1>, <2,2>} is a set of linear independent vectors but is not a space.

Answer 8

i think 2nd one is not because there is a domain for which c1 or c2 does not exist due to the sqrt

Answer 9

errr
{<1,1>, <2,3>} is a set of linear independent vectors but is not a space.

Answer 10

haha ya i was thinking about that

Answer 11

wait how come its not a vector space?

Answer 12

how do I prove it?

Answer 13

do u mean like its a vector space only if its
c1(v1)+c2(v2)

Answer 14

where v1 and v2 are linearly independant

Answer 15

so take two vectors of the form \(c_1<1,0>+c_2<0,1>\) like \(c_1<1,0>+c_2<0,1>\) and \(c_3<1,0>+c_4<0,1>\) and add them

\(c_1<1,0>+c_2<0,1>+c_3<1,0>+c_4<0,1>= <c_1,c_2>+<c_3,c_4>=\\(c_1+c_3)<1,0>+(c_2+c_4)<0,1>\) Since \(c,_1,c_2,c_3,c_4 \in \mathbb{R}\) we have that \(c_1+c_3,c_2+c_4 \in \mathbb{R}\) so \((c_1+c_3)<1,0>+(c_2+c_4)<0,1>\in A\)

Where I am calling A the set in a)

Answer 16

so we are closed with respect to vector addition right>

Answer 17

note we need only show its a subset of \(\mathbb{R}^2\) because \(\mathbb{R}^2\) is surely a vectorspace

Answer 18

yes it passes vector addition properties now what about scalar multiplication and zro vector

Answer 19

you can easily just throw in scalars on my last proof to show scalar closure

Answer 20

and zero vector? is it even in there?

Answer 21

of course \(c_1=0, c_2=0\)

Answer 22

ahhh, i see. So c1 + c2 can be anything....this concept is confusing and also makes sense at the same time

Answer 23

for scalar

\(\alpha (c_1<1,0>+c_2<0,1>)+\beta(c_3<1,0>+c_4<0,1>)= <\alpha c_1,\beta c_2>+<\alpha c_3,\beta c_4>=\\\alpha(c_1+c_3)<1,0>+\beta(c_2+c_4)<0,1>\)
where \(\alpha,\beta\) are in some field \(F\)

Answer 24

that shows both closure of addition and multiplication

Answer 25

because again \(\alpha(c_1+c_3),\beta(c_2+c_4)\in \mathbb{R}\)

Answer 26

makes sense completely. Thanks a lot

Answer 27

what about part b

Answer 28

that should say

\(\alpha (c_1<1,0>+c_2<0,1>)+\beta(c_3<1,0>+c_4<0,1>)= \\<\alpha c_1,\alpha c_2>+<\beta c_3,\beta c_4>=(\alpha c_1 + \beta c_3) <1,0>+(\alpha c_2+\beta c_4)<0,1>\)

Answer 29

um

Answer 30

sec

Answer 31

2 is not because
\(<1,0>\) is in the set, but \(<1,0>+<1,0>=<2,0>=2<1,0>+0<0,1>\notin A\)

Answer 32

it is not closed wrt vector addtion

Answer 33

sweet, thank you soooo much

Answer 34

because \(2+0\ne 1\)

Answer 35

haha ya i figured tht

Answer 36

or \(2^2+0^2 = 4 \ne 1\)

Answer 37

ok gl, must sleep.

Answer 38

haha sleep is not for math and science major good joke tho

Answer 39

starting my phd in 2 weeks, and I have only been graduated for 2 weeks... I get sleep now:)

Answer 40

congrats and good luck friend

Answer 41

tnx ttys im sure:)

Answer 42

yep me too :)

Answer 43

thanks for all help