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Let $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}, f(x)=Ax$ where $$A = \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{pmatrix}$$

Check if $f$ is injective and or surjective.

I think what I need to know to answer that question is the rank of $f$.

And then if the rank is equal to the amount of colums of the matrix, then it is injective.

If the rank equals to the amount of rows of the matrix, then it is surjective.

If rank = amount of rows = amount of colums then it's bijective.

Is this correct so far?

But the problem is I don't know how to do that for this matrice, calculating the rank :(

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    Just reduce the matrix – 2017-01-22

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f can't be injective since rank $A$ must be less than $3=$ number of columns, since it has only two rows.

Its reduced echelon form is $$ \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix},$$ so rank$A=2$. Since rank$A$ equals number of rows, then $f$ is surjective.

Apply row operations: \begin{equation*} \begin{aligned} &&\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \\ &\Rightarrow &\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \end{bmatrix}\\ &\Rightarrow &\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}\\ &\Rightarrow &\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix}. \\ \end{aligned} \end{equation*}

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    Thank you! Maybe can you shortly explain how you made the matrix smaller? – 2017-01-22
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    Do you really need row reduction to see this matrix has rank $2$? – 2017-01-22
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    @Bernard No, but I though it was more clear to the asker. – 2017-01-22
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$$\dim \ker f=3-\text{rank }A=3-2=1\ne0\Rightarrow f\text{ is not injective.}$$

$$\dim \text{Im } f=\text{rank }A=2=\dim\mathbb{R}^2\Rightarrow f\text{ is surjective.}$$