Recall that `ZZ^(n)_p` ={(`x_(1)`,...,`x_(p)` ):`x_(1)`,...,`x_(p)` `in` `ZZ^n`} is a vector space over `ZZ^n.`` `

Consider the subspace S=span{(1,2,1), (2,1,1)} of `ZZ^(3)_2`. Do the vectors (3,3,3) and (0,3,1) belong to S? Find a basis for the subspace S=span{(1,1,1,1), (1,0,1,2), (0,2,2,1)} of ` ``ZZ^(4)_3`.

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`a)`

`S` `in ZZ_5^3` is described by

` ``r([1],[2],[1]) + s([2],[1],[1]) = ([x],[y],[z])`

`(x,y,z) = (3,3,3)` does not belong to `S` :

`r=1` and `s=1` satisfies the requirement for `x` and `y`but leaves `z= 2`, ie `(x,y,z) = (3,3,2)` is in `S` but not `(3,3,3)` .

` ` `(x,y,z) = (0,3,1)` however does belong to `S`:

1 2 | 0 reduced gives 1 2 | 0

2 1 | 3 0 -3 | 3

1 1 | 1 0 -1 | 1

implying `s = -1` and hence `r = 2`.

b)

Give a basis for the subspace `S = span{(1,1,1,1),(1,0,1,2),(0,2,2,1)}` of `ZZ_3^4`

First check that `B = {([1],[1],[1],[1]),([1],[0],[1],[2]),([0],[2],[2],[1])}` (ie the elements which `S` spans). We know that `B` spans `S` already by definition. The other condition is that the elements of `B` (`v_1,v_2,v_3)` are linearly independent.

This is true if `a_1v_1 + a_2v_2 + a_3v_3 = 0 ` `implies` `a_1=a_2 = a_3 = 0`.

Now by row reduction we get

1 1 0 `->` 1 1 0 `->` 1 1 0

1 0 2 0 -1 2 0 1 -2

1 1 2 0 0 2 0 0 2

1 2 1 0 2 1 0 0 3

` `From the last two rows we see that this implies `a_3 = 0` (otherwise there is an inconsistency) which then further implies that `a_2 = a_1 = 0`.

Hence `B` is indeed a basis for `S`.

**Answer**

**a) (3,3,3) does not belong to S whereas (0,3,1) does**

**b) A basis for S is simply the vector elements which it is defined to span (ie B = {(1,1,1,1),(1,0,1,2),(0,2,2,1)}, since these are linearly independent**

NB, `S` is in `ZZ_2^3` , not `ZZ_5^3`.

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