Question

Need help! Let F = Z3, and let W = { (a1, a2, a3, a4, a5) ∈ F5 | a1 + a5 = 0, a2 + 2a4 = 0 } (so that
W is the set of all solutions to a homogeneous system of linear equations with coefficients
from Z3 and is therefore a subspace of F5). Find a basis for W, and determine dim(W).

Answer 1

basically, how do you do arithmetic in Z3? with {0,1,2}...supposedly you have to solve the system of equations but I dont know how to proceed

Answer 2

\[a_1+a_5=0\]
\[a_1=-a_5\] so choices here are
\[a_1=a_5=0\]
\[a_1=1,a_5=2\] or
\[a_5=1, a_1=2\]

Answer 3

ok, I understand that...to form a basis now, do we have to include all possible quintuples that offer solutions? because I did that and got more than 5 possibilities which cant be, needs to be dim5 or lower

Answer 4

\[a_2+2a_4=0\]
\[a_2=a_4=0\] or
\[a_1=a_4=1\] or
\[a_1=a_4=2\]

Answer 5

you are working in
\[\mathbb{Z_3}\] right?

Answer 6

yup

Answer 7

ok now i have to think...
first equation says
\[a_1=-a_5\]
second says
\[a_2=a_4\]

Answer 8

right, I took all possible scenarios and got 9 vectors, but it has to be 5 or less right? such as (0,0,0,0,0), (0,1,a3,1,0)...

Answer 9

why does it have to be less than 5?

Answer 10

you have a total of
\[5^3=125\] possible vectors in this vector space.

Answer 11

I thought the dimension has to be less than 5? because F^5

Answer 12

i mean in the entire vector space. number of 5 tuples and three choices for each entry.

Answer 13

Here is a reply I got about the question:Yes, the dimension of F^5 is 5 (basis e_1,e_2,e_3,e_4,e_5). This is true for any (every) field F.

You just need to solve the system of linear equations ( the only difference
is that the arithmetic is done in Z_3, not in R). The dimension of W is definitely less than that of F_5,
since in a finite dimensional space, every proper subspace has dimension strictly less than the dimension of
the whole space.

Answer 14

I agree with you though, that's how I think

Answer 15

oh yes the "dimension" is 5, but that does not mean there are five vectors only

Answer 16

ok true perhaps, but there could only be 5 independent vectors right?

Answer 17

we just have to come up with a basis, which should be easy if i could remember how to do this in general. the basis for the whole thing is the usual one
\[\{(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)\}\] but we need a basis for this subspace

Answer 18

right...we need to prove that basis is lin independent (i.e Ci=0 for CiVi) and that it generates W

Answer 19

i have to think for a minute, but this should be straight forward. you have to come up with linearly independent vectors that span your whole subspace. there may only be one

Answer 20

I found 9 vectors, (a3 can be anything right?) which would generate W, would the next step be to find if they are linearly independent?

Answer 21

yes, and my guess is that there may be 3

Answer 22

i mean 3 linearly independent ones

Answer 23

how would you proceed with eliminating vectors?

Answer 24

(0,0,1,0,0)
(1,0,0,0,2)
(0,1,0,1,0) is my guess

Answer 25

ok how would you get (2,0,0,0,1)

Answer 26

because as you said
\[a_3\] can be anything so we have (0,0,1,0,0) as a basis

Answer 27

i mean a basis vector

Answer 28

\[2(1,0,0,0,2)=(2,0,0,0,1)\]

Answer 29

ahhh right...ok

Answer 30

So, find all possible solutions to W, then find the lin ind set of those vectors right?

Answer 31

ok i think it is getting clearer to me, maybe. think of it this way:
suppose you were working over the real numbers instead of Z3
then the equations would give you
\[a_1=-a_5, a_2=-2a-4\] so you would have a basis that looked like
\[(1,0,0,0,-1), (0,0,1,0,0), (0,2,0,1,0)\]

Answer 32

typo there, i meant
\[a_2=-2a_4\] of course

Answer 33

yup

Answer 34

and also
\[(1,0,0,0,-1), (0,0,1,0,0), (0,2,0,-1,0)\]

Answer 35

so it is a similar situation here but much easier because you only have three numbers to choose from . so i am going to bet that the basis is the one i wrote, and you can easily check that your nine vectors are linear combinations of the three

Answer 36

i mean i would not bet a weeks pay on it, but i am pretty sure it is right

Answer 37

and you could also show that the three are linearly independent. and that would not be much

Answer 38

but it wont have -1 because that isn't an element of Z3

Answer 39

right no -1 but that is ok because -1 = 2

Answer 40

ok thanks a lot, I think I know how to proceedd...

Answer 41

(0,0,1,0,0) (1,0,0,0,2) (0,1,0,1,0)
becase a3 can be anything, a2 and a4 are equal, and whatever a1 is a5 is the other one

Answer 42

yw