# Basis and dimension of row/column space

I'm having trouble relating row/column space to basis and dimension.

Row space is the span of all linear combinations of the rows of $A$ and likewise column space is the span of all linear combinations of the columns of $A$.

So given a matrix, $$A= \begin{bmatrix} 1 & 3 & 0 & 2 & 3 \\ 2 & 1 & -1 & 1 & -1 \\ -1 & 1 & 0 & 1 & 3\\ \end{bmatrix}$$

the row space of $A$ is just $$\operatorname{span}(\langle1,3,0,2,3\rangle, \langle2,1,-1,1,-1\rangle, \langle-1,1,0,1,3\rangle)$$ (I can't figure out how to typeset span/vectors) and the column space is $$\operatorname{span}(\langle1,2,-1\rangle, \langle3,1,1\rangle, \langle0,-1,0\rangle, \langle2,1,1\rangle, \langle3,-1,3\rangle)$$

So how do I find the basis of the column space? I have to row reduce my matrix first right? Which gives me:

$$\begin{bmatrix} 1 & 0 & 0 & -1/4 & -3/2 \\ 0 & 1 & 0 & 3/4 & 3/2 \\ 0 & 0 & 1 & -3/4 & -1/2\\ \end{bmatrix}$$

So wherever I have a pivot in my rref matrix, whichever column that corresponds to in the original matrix forms my basis of the column space. What about the row basis? Is it then just vectors from the rows of the rref?

And dimensions, is the dimension of my column space equivalent to the number of columns with a pivot? So is this the same as the dimension of the null space? And what about the dimension of the row space.

• if the rows are linearly independent then they are a basis of this space, the same for the column space. So you only need to reduce the list of vectors, of rows or columns, to a list of linearly independent vectors. Commented Jul 24, 2017 at 11:07

$$rank(A)+ nullity(A) = \text{number of columns}$$