# Basis for the row space, column space and null space of a matrix

Let $A$ = \begin{pmatrix}11&-2&36&2\\ -2&1&-4&0\\ 3&0&12&1\\ 1&-1&0&0\end{pmatrix}

Determine a basis for the row space, column space and null space."

So I know the dimension of the row and column space is 3 and the dimension of the null space is 1. But I'm not entirely sure how to proceed further to determine the basis. For the row and column space, am I supposed to find a set of row and column vectors respectively that have a non-zero determinant (Linearly independent)? And how exactly am I supposed to find the basis for the null space?

Any help?

• Look at "related" suggestions on the right hand side of the page. Try following this method for a start. Commented Nov 29, 2017 at 0:03
• Please, if you are ok, you can accept the answer and set it as solved. Thanks!
– user
Commented Jan 10, 2018 at 22:19

This matrix reduces, through row reduction, to $\begin{pmatrix}1 & -1 & 0 & 0 \\ 0 & -1 & -4 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The row space has the three basis vectors, (1, -1, 0, 0), (0, -1, -4, 0), and (0, 0, 0, 1), so dimension 3. The column space has the three basis vectors (1, 0, 0, 0), (-1, -1, 0, 0), and (0, 0, 1, 0). (0, -4, 0, 0) is not independent because (0, -4, 0, 0)= 4(-1, -1, 0, 0)+ 4(1, 0, 0, 0). The column space has dimension 3. That's always true- the dimension of the row space of a matrix is equal to the dimension of the column space".

(x, y, z, t) is in the "null space" if and only if $\begin{pmatrix}11 & -2 & 36 & 2 \\ -2 & 1 & -4 & 0 \\ 3 & 0 & 12 & 1 \\ 1 & -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ t\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix}$. That is equivalent to the four equations 11x- 2y+ 36z+ 2t= 0, -2x+ y- 4z= 0, 3x+ 12z+ t= 0, x- y= 0. From x- y= 0, of course, y= x so the other three equations can be written 9x+ 36z+ 2t= 0, -x- 4z= 0, and 3x+ 12z+ t= 0. From -x- 4z= 0, x= -4z so the other two equations can be written 2t= 0, 0= 0, and t= 0. Clearly t= 0 but we cannot solve for numerical values of x, y, and z. We can say that (x, y, z, t)= (-4z, -4z, z)= z(-4, -4, 1) where z can be any number. That is a basis for the null space is {(-4, -4, 1)} and the dimension of the null space (the "nullity") is 1.

Note that the dimension of the null space, 1, plus the dimension of the row space, 1+ 3= 4, the dimension of the whole space. That is always true. After finding a basis for the row space, by row reduction, so that its dimension was 3, we could have immediately said that the column space had the same dimension, 3, and that the dimension of the null space was 4- 3= 1 without any more computation.

• The basis that you gave for Column space is incorrect. How will it generate a vector with non zero last entry with your basis. Commented Jan 30, 2022 at 15:52

yes for the row and column space you have to find a set of row and column vectors respectively that are Linearly independent.

For the null space here is an example: Finding the basis of a null space

• Yes, I was pretty sure about that. Thanks for confirming! How about the basis for the null space though? Commented Nov 29, 2017 at 0:01
• If (as the OP says) the dimension is $3$ then you will be looking at a $3\times4$ or $4\times3$ matrix. Such a matrix does not have a determinant. Commented Nov 29, 2017 at 0:01
• @David thanks I've corrected my typo
– user
Commented Nov 29, 2017 at 0:03