# 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.

• The nonzero rows will give you a basis for the row space, not the nullspace. To find a basis for the nullspace, you want to solve $Ax=0$ and then find a basis for this solution space. – user84413 Aug 26 '13 at 20:24
• oh yes, sorry, i'll correct it – Aman Mittal Aug 26 '13 at 20:24
• @user84413 ok, can you please elaborate on how that works ? – Aman Mittal Aug 26 '13 at 20:25
• You want to augment A with a column of zeros at the end, and then reduce this augmented matrix to echelon form. Then let the variables which do not corresponding to leading 1's be arbitrary parameters (r,s,t, etc), and solve for the variables corresponding to the leading 1's in terms of these parameters. – user84413 Aug 26 '13 at 20:33

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.