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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 ?

2 Answers 2

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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.

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    So how do i generally proceed to find the image ? I know that the coloumns in the matrix are linear dependent.3rd coloumn is linear combination of first and second. Can i remove for example the third coloumn and say that only first two coloumns span the image ? – 2017-01-26
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    Exactly! Your image is the span of the matrix's columns, and one column is *already in* the span of the other two. So you don't need the final one; the span is unchanged removing it. It does have rank 2, so you can't remove anything else. – 2017-01-26
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    So in this manner i could show one inclusion ? How could i show the other inclusion ? – 2017-01-26
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    To make sure I understand: Let's call your columns $v_1, v_2$, and $v_3$. You're saying, how to show $\langle v_1, v_2, v_3 \rangle = \langle v_1, v_2 \rangle$? Well, the $\supseteq$ direction is clear; the $\subseteq$ direction is because we can write $a_1 v_1 + a_2 v_2 + a_3 v_3$ as $(a_1 + a_3)v_1 + (a_2 - \frac{1}{3}a_3)v_2$, since $v_3 = v_1 - \frac{1}{3}v_2$. So $v_1$ and $v_2$ suffice to generate every linear combination of $v_1, v_2, v_3$. – 2017-01-26
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    o no i was actually asking about the inclusion with image f so is included in image of f, how could i show the other inclusion ? – 2017-01-26
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    Think more about what I've said so far, in the answer and in that latest comment. If you're still not sure in a couple of hours, I can expand. – 2017-01-26
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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 $$\left\{\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}0\\6\\7\end{bmatrix}\right\}.$$