Non-zero rows give the basis of null space? Q. Find the nullity and basis of null space of linear transformation $A:R^4\rightarrow R^4$ given by the matrix
$$\begin{bmatrix} 0 & 1 & -3 & -1 \\ 1 & 0 & 1 & 1 \\ 3 & 1 & 0 & 2 \\ 1 & 1 &-2&0 
\end{bmatrix}$$
Its given that if we reduce this matrix to echlon form then the non-zero rows gives the basis of null space. I do not understand why this works.
 A: The basic principal is that elementary row options do not change the rowspace of a matrix. To prove that, you could demonstrate that the processes of scaling a basis element, interchanging two basis elements, and adding multiples of one basis element to another, all result in a new basis.
So, if you perform elementary row operations and change it entirely to row-echelon form, you are left with a matrix that has some zero rows and some nonzero rows. The row-echelon shape makes it obvious that the nonzero rows are linearly independent. Can you see why?
Since this is the case, the remaining rows are a basis for the rowspace of the matrix.
Basic row operations do not change the (right) nullspace either, so you can use the new matrix to work with the nullspace. Using that matrix instead of the original matrix, it will be particularly easy to solve the equation $A'x=0$ with back substitution to discover what is in the nullspace.
A: Basic operations on rows do not change the null space. When you add for example two rows, it's like adding two equations in the system obtained when writing Ax=b.
Edit : the first answer is more complete, didn't see it when writing mine.
