Column Space/Row Space I just have a small question. I was wondering if someone could explain to me the difference between "column space" and "basis for column space" as well as "row space" and "basis for row space".
I've looked everywhere for an explanation but all I get is what a basis for a column/row space is.
 A: Given an $m \times n$ matrix $A = (a_{ij})$, the column space of $A$ is the subspace of $K^m$ spanned by the columns of $A$ (which are vectors in $K^m$ since they have $m$ entries).
Clearly, columns need not to be linearly independent. For example if $A$ is
$$A= \left(\begin{matrix}6 & 8&0 \\ 3& 4& 0\end{matrix} \right)$$
you can easily see that $A$ has 3 columns (belonging to $K^2$) so the column space of $A$ is the space spanned by these 3 vectors. Let's call
$$C= \left< \left(\begin{matrix}6  \\3\end{matrix} \right), \left(\begin{matrix}8  \\4\end{matrix} \right), \left(\begin{matrix}0  \\0\end{matrix} \right)\right>$$
Now, $C$ has dimension $1$ (check). So you can say that the column space of $A$ is the subspace of $K^2$ spanned by the first column.
So a basis for $C$ is $\left(\begin{matrix}6  \\3\end{matrix} \right)$.
A: It's the same as with any basis and vector space.
The rows of a matrix span the row space. Let's say your matrix had rows ${\bf r}_1,\ldots,{\bf r}_n$, then the row space is every vector of the form
$\alpha_1{\bf r}_1+\cdots+\alpha_n{\bf r}_n$, where the $\alpha_i$ are numbers.
The rows ${\bf r}_1,\ldots,{\bf r}_n$ might not form a basis for the row space because they might not be linearly independent. To find a basis for the row space, put your matrix into row echelon form. The non-zero rows of the matrix will then give a basis for the row space.
The same is true for the column space. You just need to put the matrix into column echelon form.
Example: Find a basis for the row space of the matrix
$$M = \pmatrix{1 & 2 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 1 \\ 3 & 1 & 6}$$
$$\pmatrix{1 & 2 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 1 \\ 3 & 1 & 6} 
\xrightarrow{R_2-2R_1}
\pmatrix{1 & 2 & 3 \\ 0 & 0 & -5 \\ 1 & 0 & 1 \\ 3 & 1 & 6} 
\xrightarrow{R_3-R_1}
\pmatrix{1 & 2 & 3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \\ 3 & 1 & 6}
\xrightarrow{R_4-3R_1}
\pmatrix{1 & 2 & 3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \\ 0 & -5 & -3} \\ \\
\xrightarrow{R_2 \to R_2/5}
\pmatrix{1 & 2 & 3 \\ 0 & 0 & -1 \\ 0 & 0 & -2 \\ 0 & -5 & -3}
\xrightarrow{R_3 - 2R_2}
\pmatrix{1 & 2 & 3 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & -5 & -3}$$
This is as good as we can do. A basis for the row space is
$\{(1,2,3),(0,0,-1),(0,-5,-3)\}$. Of course, we can tidy this up a bit any used $\{(1,2,3),(0,0,1),(0,5,3)\}$ instead. The row space is
$$\{ \alpha (1,2,3)+\beta(0,0,1)+\gamma(0,5,3) : \alpha,\beta,\gamma \in \mathbb{K} \} \cong \mathbb{K}^3$$
A: Similarly, the "row space" of a matrix is the subspace spanned by the rows!  Using Crostul's example, $$\begin{bmatrix}6 & 8 & 0 \\ 3 & 4 & 0 \end{bmatrix},$$ the row space is spanned by the two vectors $(6, 8, 0)$ and $(3, 4, 0)$ but it is obvious that those are dependent so that same row space is spanned by $(3, 4, 0)$ alone (and, geometrically, is the line $y= \frac{4}{3}x $in the $xy$-plane).
Those are the definitions of "column space" and "row space".  Of course, since the given matrix has 3 columns and 2 rows, it clearly maps vectors in R3 to R2.  If we were to apply the matrix to the easy basis vectors, $(1, 0, 0)^T, (0, 1, 0)^T$, and $(0, 0, 1)^T$ we would get precisely $(6, 3)^T, (8, 4)^t$, and $(0, 0)^T$ so that the column space is the "range" of the matrix.  Further, if we were to apply this matrix to $(-4, 3, 0)^T$ we would get $$\begin{bmatrix}6 & 8 & 0 \\ 3 & 4 & 0 \end{bmatrix}\cdot\begin{bmatrix}-4 \\ 3 \\ 0 \end{bmatrix}= \begin{bmatrix}6(-4)+ 3(4)+ 0(0) \\ 3(-4)+ 4(3)+ 0(0)\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}.$$  Further applying it to $(0, 0, 1)^T$ would give $$\begin{bmatrix}6 & 8 & 0 \\ 3 & 4 & 0 \end{bmatrix}\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}6(0)+ 8(0)+ 0(1) \\ 3(0)+ 4(0)+ 0(1)\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}.$$
I chose those vectors because they are orthogonal to (3, 4, 0).  That is, the row space is the orthogonal complement to the null space of the matrix.
