# Is the map linear?

$$\begin{bmatrix}a&b\\\\c&d\end{bmatrix}\mapsto \begin{bmatrix}ad-bc\\\\0\\\\0\end{bmatrix}$$

If it is linear I need to find a basis for the kernel and image but I am struggling to do this so i don't think it's linear but I have no idea why.

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It's not linear because:

$\begin{bmatrix}1&0\\\\0&0\end{bmatrix}$$\mapsto \begin{bmatrix}0\\\\0\\\\0\end{bmatrix} \begin{bmatrix}0&0\\\\0&1\end{bmatrix}$$\mapsto$ $\begin{bmatrix}0\\\\0\\\\0\end{bmatrix}$

$\begin{bmatrix}1&0\\\\0&1\end{bmatrix}$$\mapsto$ $\begin{bmatrix}1\\\\0\\\\0\end{bmatrix}\neq \begin{bmatrix}0\\\\0\\\\0\end{bmatrix}+\begin{bmatrix}0\\\\0\\\\0\end{bmatrix}$

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I undeleted this because it is a correct answer. If you deleted it for another reason, let me know. –  robjohn Dec 5 '12 at 16:46
Let $X$ be a $2\times 2$ matrix. Call your map $F(X)$. Is $F(2X) = 2F(X)$?
so it is quite easy to see the kernel of $F$, you just have to look for $M \in M_2$ so that $det(M)=0$ ...