I dont know if this is a stupid question or not but i wondering is they a difference between a question asking you to find the basis for the null space of A and determining the null space of A
Determine the Null space of $A$ vs Finding the basis for the Null Space of $A$
linear algebramatricesvector-spaces
Related Solutions
The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$. To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.
Suppose after doing this, you obtain $$ \left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr 0&0&0&0&0 \cr }\right] $$
Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$ Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.
The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.
So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.
Now solve for $x_1$ and $x_3$:
The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.
So, the general solution to $A{\bf x}={\bf 0}$ is $$ {\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right] $$
Let's pause for a second. We know:
1) The null space of $A$ consists of all vectors of the form $\bf x $ above.
2) The dimension of the null space is 3.
3) We need three independent vectors for our basis for the null space.
So what we can do is take $\bf x$ and split it up as follows:
$$\eqalign{ {\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right] &=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+ \left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+ \left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr &= a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+ c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+ b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr } $$ Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.
I'm not sure that this answers your question. I did a bit of "hand waving" here. What I glossed over were the facts:
1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).
2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).
Your particular example has a simple answer. The subspace whose basis is the rows of your matrix is called a first-order Reed-Muller code of length 16, and its dual (null space) is the second-order Reed-Muller code of length 16. Denoting the rows by $1, x_1, x_2, x_3, x_4$ respectively, the dual code has basis vectors that are $$1, x_1, x_2, x_3, x_4, x_1x_2, x_1x_3, x_1x_4, x_2x_3, x_2x_4, x_3x_4$$ where $x_ix_j$ is the element-by-element product of the row vectors, e.g. $x_1x_2=0000000000001111$ and $x_2x_3=0000001100000011$. More generally, the dual of the first-order Reed-Muller code of length $2^m$ is the $(m-2)^{\text{th}}$-order Reed-Muller code of length $2^m$, also known as the extended Hamming code of length $2^m$, and the basis vectors can be taken as all the monomials of degree $m-2$ or less.
More generally, in $\mathbb F_2^n$, given a $k\times n$ matrix of row rank $k$, express it in row-echelon form, interchange columns and rows as needed to express the matrix in the form $[I_{k\times k}\quad P]$ where $P$ is a ${k\times(n-k)}$ matrix. The null space is spanned by $[P^T\quad I_{(n-k)\times(n-k)}]$. Now undo the column permutations to get the basis vectors for the original problem. For vector spaces $\mathbb F_q^n$ where $q$ is not a power of $2$, use $-P^T$ instead of $P^T$.
Best Answer
I am assuming $A$ is a real $n\times n$ matrix.
Null space of $A$, say $N(A)$ is the set of $x\in \mathbb{R}^n$ such that $Ax=0$. It is an easy exercise to verify that it a subspace of $\mathbb{R}^n$.
Now every vector space has a basis and the basis completely determines it. As $N(A)$ is a vector space, it is completely specified by it's basis.
If you don't get a basis for $N(A)$, still $N(A)$ can be determined if you have a spanning set for $N(A)$ (basis would be some subset of it).