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.
There is something about this not clicking for me, any clarification is appreciated.
Best Answer
For the row basis, the non-zero rows in the RREF forms the basis. This is due to elementary row operations does not change the row space and also the non-zero rows are linearly independent.
Dimension of column space is equal to the number of columns with a pivot. It is known that the dimension of row space is equal to the dimension of column space. To see this, just notice that the number of pivot is equal to number of non-zero rows in the RREF.
You might find rank-nullity theorem helpful, that is
$$rank(A)+ nullity(A) = \text{number of columns}$$
Nullity is the dimension of null space.