Question

we have a vector sum on R^2 and define it as: (x1,y1)+(x2,y2)=(x1+x2,y1,y2). Explain why these scalar products aren't a good choice to make R^2 a vectorspace

Answer 1

a) a(x,y)=(x,y)
b) a(x,y)=(ax,0)

Answer 2

my friend has this as his answer:
\[1.v \neq v\]

Answer 3

and
\[1.(u_1,u_2)=(u_1,0) \neq (u_1,u_2)\]

Answer 4

how did he get this?

Answer 5

is this a typo
(x1,y1)+(x2,y2)=(x1+x2,y1,y2)
?

as written, your result is not in R^2

Answer 6

odd..not a typo...

Answer 7

i'm guessing he meant: (x1,y1)+(x2,y2)=(x1+x2,y1+y2)

Answer 8

@phi

Answer 9

@amistre64

Answer 10

please someone explain this..

Answer 11

what do a and b mean? are they just options to consider?

Answer 12

a(x,y) = (x,y) for some any a in R does not provide for an inverse element of (x,y)

Answer 13

and the other doesnt appear to provide for an identity

Answer 14

other than that ... thats all i can make of it

Answer 15

i think its just question a) and b)

Answer 16

then what i wrote prolly pertains .... its not that condusive to define the scalar multiplication in that manner since it would mess with identity and inverse porperties of vector spaces

Answer 17

the answers my friend wrote, how did he come up with it? and what does he even mean with v?

Answer 18

if you are asking questions about what your friend came up with ... you would prolly get better results by asking your friend.

Answer 19

true, but i can only contact him next week

Answer 20

his first result is off tho; 0(a,b) not= (0,0)

Answer 21

alright..