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Just a simple question, but I had a hard time finding a decent explanation. I'm confused about finding the image when given a matrix and what exactly this represents. For example the matrix A = \begin{bmatrix}1&2&3\\1&3&4\\1&4&5\end{bmatrix} What is the image of $F_A$?

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    The image is $\{(x+2y+3z,x+3y+4z,x+4y+5z):x,y,z\in \mathbb{R}\}$ – 2017-02-10

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The image is the set of all vectors of the form $Av$ for an arbitrary vector $v$.

For example for the standard basis vector $e_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$, we know $Ae_1=\begin{bmatrix}1\\1\\1\end{bmatrix}$ is in the image.

Hint 1: what happens for the other standard basis vectors $e_2$ and $e_3$?

Hint 2: if you know $Ae_1$, $Ae_2$, and $Ae_3$, then for any $v = c_1e_1+c_2e_2+c_3e_3$ you can write $Av = c_1(Ae_1)+c_2(Ae_2)+c_3(Ae_3)$. Can you use this to describe the image of $A$ neatly?

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The image of a matrix, or of the linear map it represents, is the same as the column space. It's easy to see the third column is the sum of the first two, and these two are linearly independ, then

$$\text{Im}\,F_A=\text{Col}_A=\text{Span}\,\left\{\;\begin{pmatrix}1\\1\\1\end{pmatrix}\;,\;\;\begin{pmatrix}2\\3\\4\end{pmatrix}\;\right\}$$$${}$$

This means a vector $\;\begin{pmatrix}x\\y\\z\end{pmatrix}\in\Bbb R^3\;$ is in $\;\text{Im}\,F_A\iff F_A\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}\;$ for some $${}$$$\;\begin{pmatrix}a\\b\\c\end{pmatrix}\in\Bbb R^3\iff \begin{pmatrix}x\\y\\z\end{pmatrix}\;$ is a linear combination of the above two vectors i nthe forst two columns of $\;A\;$ ..

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    So is there a simple way to know whether a vector is in the image? – 2017-02-10
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    @JanoyCresva YEs: check it belongs to the column space...you can form a new matrix with the lin. independent columns and below them the vector you want to check, and begin reducing: if you get the last row becomes all zeros then the vector belongs to the image, otherwise it doesn't. – 2017-02-11
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Your matrix $A$ represents a linear transformation, i.e. $F_A:\mathbb{R}^3 \to \mathbb{R}^3$, so the image is $x \mapsto Ax$.