Question

I am trying to understand the whole idea of finding zero vectors, How do you find them?

Answer 1

row reduction ....

Answer 2

can you be more specific? im assuming this is a linear algebra concept, but i cant be sure

Answer 3

yes, its linear algebra, but in vector spaces

Answer 4

do you have a problem that can be used as an example?

Answer 5

ok, let me type it now

Answer 6

Let V = { * }, with vector addition +: V x V -->V defined by * +* = * and k* =* for k is an element of real numbers R. Check that the 10 axioms for vector spaces hold, and find the zero vector. Find -*

Answer 7

please treat * as a big star

Answer 8

so, it looks like the elements of the set consist of 1 thing: *

addition (or the operation on the set) is such that *+*=*, so we have closure. And it seems like a defined property is given also as k* = * for some arbitrary real constant scalar.

Answer 9

yes, it is closed under addition

Answer 10

hmm, well the operation is also defined as its identity: e+* = * when e=*

inverse; * + (-*) = e
* + (-*) = *
so it is also its own inverse ....

i do not see a way to generally define Ax = 0 since k*=* and there is no 0 to be had.

Answer 11

I don't know what to say man because I have no clue man

Answer 12

does it say that this IS a vector space? or are you trying to prove that it is or is not a vector space?

Answer 13

ok, i want u to help me may be with an example, on how to find zero vector of space, is it doable?

Answer 14

depends on how much a recall :) but i can try

Answer 15

please

Answer 16

the zero vector can be determined, given a rule defining an operation, if the operation will produce 0

Answer 17

if we define a+a = a, and ka = a

then there is no way to really express a zero vector unless we assume a=0 to start with.

Answer 18

and say a.0 =0+0 =0, that's where it hits me the most. like does the 0+0 come from

Answer 19

it might help if you refresh my memory as to the 10 axioms of a vector space. I recall some things equating inverse with zero vectors, but its a bit fuzzy at the moment

Answer 20

A real vector space
is a set X with a special element 0, and three operations:

Addition: Given two elements x, y in X, one can form the sum x+y, which is also an element of X.

Inverse: Given an element x in X, one can form the inverse -x, which is also an element of X.

Scalar multiplication: Given an element x in X and a real number c, one can form the product cx, which is also an element of X.

These operations must satisfy the following axioms:
Additive axioms.
For every x,y,z in X, we have x+y = y+x. (commutative)
(x+y)+z = x+(y+z). (associative)
0+x = x+0 = x. (identity)
(-x) + x = x + (-x) = 0. (inverse)

Multiplicative axioms.
For every x in X and real numbers c,d, we have
0x = 0 (zero vector)
1x = x (identity)
(cd)x = c(dx) (assiciative)

Distributive axioms.
For every x,y in X and real numbers c,d, we have
c(x+y) = cx + cy.
(c+d)x = cx +dx.

Answer 21

http://www.math.uconn.edu/~troby/Math2210F13/LT/sec4_1.pdf

here is another site that simply says: there is an element (the zero vector)
such that * + z = *; in this case i would then say the zero vector, z=*

Answer 22

yes, it is the zero vector that I seek to understand.How do you apply it inorder to find zero vector?

Answer 23

for vector addition, the "zero vector" is the identity element. it just seems odd for a 1 element set.

Answer 24

if we simply rename the information as:

Let V = { 0 }, with vector addition +: V x V -->V defined by
0 +0 = 0
and k0 =0 for k in R.

Check that the 10 axioms for vector spaces hold, and find the zero vector. Find -*

it is isomorphic to the original information and easier to "see"

Answer 25

yeah. this stuff requires wide reading to get the head and the tail, but i really appreciate the website. I am checking it now. what u r saying now adds more sense to the whole qquestion.

Answer 26

1, closure; 0+0 = 0 is good

2, commutative; 0 + 0 = 0 + 0 is good

3, associative; 0 + (0 + 0) = 0 + 0 = 0
(0 + 0) + 0 = 0 + 0 = 0 is good

4, there is an e element in V such that 0 + e = 0
e = 0 is an element in V

5, inverse; there is an element i in V such that 0 + i = 0
i = 0 is an element in V

6, closure again k0 = 0 ... is good

7 and 8 are distributive laws: k(0+0) = k0 + k0 = 0 + 0, are fine

9 is associative again, c(k0) = c0 = 0
(ck)0 = 0 since ck is just some real value

10 indentity: 1(0) = 0 seems good

Answer 27

if we change all the 0s in that to *s, all the properties still hold. It is just simpler to "see" with the element being a 0 :)

Answer 28

thanx @amistre64 . really appreciate this

Answer 29

youre welcome.