# Math Help - Dimension

1. ## Dimension

what is the dimension of the following subspace:
the space of all n-sized vectors in the binary field with an even number of 1s.

could not think of a good method to do this

2. you are really expected to find all vectors the sum of whose entries is 0. so that gives you just one linear equation. that is the meaning of even number of terms in this case.

what is the dimension of its solution space ?? n ? n-1 ? n-2 ?

3. so say there is some vector u in this space, the linear equation we are looking for is u(1)+u(2)+u(3)+...+u(n-1)+u(n)=0. i dont understand ur next line where u say what is the dimension of its solution space though.

4. 1. yes you are correct.
2. i am asking you to work out the dimension of the space of vectors which satisfy this linear equation.

first you need to convince urself that it is a space. use any means to work out its dimension. try to generalize the result.

OK let me give some hints ... can you construct a matrix from that equation ? what can you say about the rank of that matrix ? is the solution space the null space of that matrix ? is rank nullity theorem applicable here ?

5. i see where u are heading with this, expecially the rank nullity theorem. i will think about forming such a matrix. thanks for the hints

6. can u give me some more tips on how to form that matrix because im not sure how i would do that.

7. OK. as u said let $u = (u_1, u_ 2, ... u_n)$ be a vector in the subspace. then $u_1 + u_ 2 + ... + u_n = 0$, which is the same as this matrix equation :
$\left[
\begin{array}{cccc}
1&1&...&1\\
0&0&...&0\\
\ldots&0&...&0\\
0&0&...&0
\end{array}
\right] \begin{pmatrix}
u_1\\
u_2\\
\ldots\\
u_n
\end{pmatrix} = 0$

i am sure now u can do it.

8. since you have come this far, i will give you some food for thought.

suppose you have m such linear equations what can you say about the matrix ? what if only r of the equations are linearly independent ? what if the linear equations are not homogeneous (meaning they are not of the form Ax=0, but looks like Ax=b) ?

in the last question, you can see clearly that solution space is not a vector space, but still can you use the techniques learnt so far to say something about it ?

the answer to all the above questions are the chief pedagogical motivations of linear algebra. the real world and historically relevant motivations come from differential equations.

9. yeah i previously tried forming the matrices in special cases n = 2,4 and 6. i tried to do a generalisation of the rank and nullity of the matrices and couldnt find any relation. i do understand though where u are going with this having the rank of the reduced matrix being the dimension of the solution space.

10. no no ... not rank, but nullity ... think again !!!

11. hmm i thought that if i have all the possible combinations that satisfy the linear equation and put them into the rows of the matrix and row reduce it, the remaining rows would be linearly independent hence being the basis of the space. the number of vectors in this basis is the dimension. correct me if im wrong.

12. yeah it is the basis of some space but not the space you wanted. it is the basis of range space. the vectors v that satisfy Av=0 belong to the kernal of A, also called null space whose dimension is called nullity.

by rank nullity theorem, rank + nullity = total dim

since you have already computed rank, you can sompute nullity by the above eqn

13. im sorry, i seem to have misunderstood u when u said form a matrix A. i think the matrix A that i have been trying to form is different from the matrix A u are talking about. can u explain what the matrix A is? from my understanding the matrix A is a matrix whose rows are all the possible vectors in the space. is that what u meant or are u referring to a different A?

14. given any system of linear equation in n variables you can write the same equation as a matrix equation of the form Av=b, where v is the vector composed of all the variables and b is a fixed vector. if b = 0 we call such system homogeneous.

i earlier wrote down a matrix equation to represent the linear equation you wanted to solve. in that equation b = 0. i am calling that matrix as A.
$
\left[
\begin{array}{cccc}
1&1&...&1\\
0&0&...&0\\
\ldots&0&...&0\\
0&0&...&0
\end{array}
\right] \begin{pmatrix}
u_1\\
u_2\\
\ldots\\
u_n
\end{pmatrix} = 0
$

so A = $
\left[
\begin{array}{cccc}
1&1&...&1\\
0&0&...&0\\
\ldots&0&...&0\\
0&0&...&0
\end{array}
\right]
$

15. oh right i see what u are doing now.... so rank(A) = 1 and we know the dimension of all vectors is n hence null(A) = n-1 and null(A) is the dimension of the space of vectors that satisfy that linear equation.

thanks for the help

i also realised that n has to be even though because if u had a vector [1 1 0] and u multiply it with 1, u get [0 0 1] which is no longer in the space, therefore when n is odd, it is not a subspace because it is not closed under scalar multiplication.