Linear algebra (image) My linear algebra is a bit rusty.
If I have a matrix, for example: $$\begin{bmatrix}2&-1&3&2\\3&-1&2&4\\1&0&-1&2\end{bmatrix}$$What is the image of this?.
Am I right to say that the image is the set of linear combinations of the columns?
Thanks!
 A: In my opinion, talking about the image of a matrix is not good terminology. An $m\times n$ matrix (over the reals) can be interpreted as a linear map $\mathbb{R}^n\to\mathbb{R}^m$ when bases of the two spaces are chosen.
One could extend this by saying that, when no bases are specified, the standard ones are chosen. But it's an abuse of language which makes things difficult when matrices associated to a linear map are considered.
On the other hand, the concept of “column space” is well defined and intrinsically linked to the matrix. It is the set of linear combinations of the columns (and it is the same as the “image” I was talking about before).
How do you find a basis for it? Just do Gaussian elimination:
\begin{align}
\begin{bmatrix}
2 & -1 &  3 & 2 \\
3 & -1 &  2 & 4 \\
1 &  0 & -1 & 2
\end{bmatrix}
&\to
\begin{bmatrix}
1 & -1/2 &  3/2 & 1 \\
3 & -1 &  2 & 4 \\
1 &  0 & -1 & 2
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & -1/2 &  3/2 & 1 \\
0 & 1/2 & -5/2 & 1 \\
0 &  1/2 & -5/2 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & -1/2 &  3/2 & 1 \\
0 & 1 & -5 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
It can be shown that the columns where the pivots are found (in this case the first and the second) form a basis for the column space. So a basis for the column space of your matrix is
$$
\left\{\;\begin{bmatrix}2\\3\\1\end{bmatrix}\,,\,
\begin{bmatrix}-1\\-1\\0\end{bmatrix}\;\right\}
$$
If you go a step further in Gaussian elimination, that is do also backwards elimination
$$
\begin{bmatrix}
1 & -1/2 &  3/2 & 1 \\
0 & 1 & -5 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & -1 & 2 \\
0 & 1 & -5 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
you can read directly the coefficients needed to express the third and fourth columns in terms of the first and the second:
\begin{gather}
\begin{bmatrix}3\\2\\-1\end{bmatrix}=
(-1)\begin{bmatrix}2\\3\\1\end{bmatrix}+(-5)
\begin{bmatrix}-1\\-1\\0\end{bmatrix}
\\
\begin{bmatrix}2\\4\\2\end{bmatrix}=
2\begin{bmatrix}2\\3\\1\end{bmatrix}+2
\begin{bmatrix}-1\\-1\\0\end{bmatrix}
\end{gather}
