Given a matrix ,how can i find the image under the linear map represented by a matrix? So if i have a matrix $\begin{pmatrix}
        1 & 9 & -2 \\
        2 & 0 & 2 \\
        3 & 6 & 1 \\
        \end{pmatrix} $
How can i find the image of this matrix ?
I tried reducing to row echelon form
\begin{pmatrix}
        1 & 0 & 1 \\
        0 & 1 & -1/3 \\
        0 & 0 & 0 \\
        \end{pmatrix}
So now the image would be something like $\langle(1,1,0)^T, (1,-3,0)^T\rangle$ am i right ?
 A: The image is  spanned by the colmun vectors of the matrix. To find a basis, you can transpose the matrix and operate row-reduction:
$$\begin{bmatrix}1&2&3\\9&0&6\\-2&2&1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&2&3\\0&-18&-21\\0&6&7\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&2&3\\0&6&7\\0&6&7\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&2&3\\0&6&7\\0&0&0\end{bmatrix}$$
Thus a basis is
$$\left\{\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}0\\6\\7\end{bmatrix}\right\}.$$
A: Let me give a two-dimensional example. 
Say we have $T\colon \Bbb R^2 \to \Bbb R^2$ given by the matrix $M_T = \pmatrix{2 & -1 \\ 6 & -3}$, so that 
$$T(x, y) = M_T\pmatrix{x \\ y} = \pmatrix{2 & -1 \\ 6 & -3}\pmatrix{x \\ y} = \pmatrix{2x - y \\ 6x - 3y} = \pmatrix{2 \\ 6}x + \pmatrix{-1 \\ -3}y.$$
Thus, by playing around with the algebra a bit, we can see that the image of our matrix is the span of its columns, $\pmatrix{2 \\ 6}$ and $\pmatrix{-1 \\ -3}$. This will always be the case.
Row-reducing the corresponding matrix doesn't in general preserve the image of the linear map. We can even see this from your example:
$$\pmatrix{1 & 9 & -2 \\ 2 & 0 & 2 \\ 3 & 6 & 1} \pmatrix{1 \\1 \\1} = \pmatrix{8 \\ 4 \\ 10},$$ but $(8, 4, 10)^T$ certainly isn't in the span of $(1, 1, 0)^T$ and $(1, -3, 0)^T$, even though it's in the image of your linear map.
