Question

Is the following set a vector space?

Answer 1

Is it closed under addition? scalar multiplication?

Answer 2

I think yes it is.

Answer 3

What if you multiply one of the vectors by \(\dfrac{1}{2}\)?
\[\frac{1}{2}\begin{pmatrix}6\\0\end{pmatrix}=\begin{pmatrix}3\\0\end{pmatrix}\]
But this vector isn't in the set, so it's not closed under scalar multiplication.

Answer 4

oh I see, but is it still closed under addition?

Answer 5

To be considered a vector space, both closure conditions must be met, so you're done as far as answering the question goes.

But no, it's also not closed under addition. Add any two vectors you like. The sum won't be in the set, aside from \(\dbinom10+\dbinom10=\dbinom20\).

Answer 6

Makes sense. Thank you. One last question - can you relate the additive identity to this question. I am confused with that too.

Answer 7

There's no zero vector in the set, so no additive identity exists.

Answer 8

sorry one more question, so if any set does not have zero vector, it is not vector space. Correct?

Answer 9

Right, that's another qualifier for a set to be considered a vector space. It comes with the set being closed under addition/multiplication, which is why these are the necessary and sufficient conditions.

Answer 10

Thanks a lot. That makes much more sense

Answer 11

yw

Answer 12

you need to have all the multiples of \(\large \begin{pmatrix}1\\0\end{pmatrix} \) to be considered as a vector space

Answer 13

so from (1 0) (2 0) ....(infinity 0)?

Answer 14

for example below is a vectorspace \[\large c\begin{pmatrix}1\\0\end{pmatrix} \]

Answer 15

c can be any real number

Answer 16

including 0

Answer 17

basically all the linear combinations of existing vectors must exist within the defined vector space

Answer 18

ahhhh. thank you. But it is not 1 of the 8 properties in my textbook. Where did you get it from?

Answer 19

yes all those dozen properties must be satisfied :)

Answer 20

the most important+useful property to be memorized is about the linear combinations

Answer 21

below is another example of vector space with dimension 2 :
\[\large c_1\begin{pmatrix}1\\0\end{pmatrix} + c_2 \begin{pmatrix} 0\\1\end{pmatrix} \]

Answer 22

never mind it is in definition....sorry about that

Answer 23

thats a vector space because has a zero vector when \(\large c_1 = c_2 = 0\), and it satisfies all the other properties described in ur book

Answer 24

so c1 and c2 can be anything we want. right?

Answer 25

they need to take ALL real values for it to be a vector space

Answer 26

it forms a FULL plane

Answer 27

youc an reach all the points in that plane by choosing c1 and c2 values