$$\begin{bmatrix}a&b\\\\c&d\end{bmatrix}\mapsto \begin{bmatrix}ad-bc\\\\0\\\\0\end{bmatrix}$$
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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$$\begin{bmatrix}a&b\\\\c&d\end{bmatrix}\mapsto \begin{bmatrix}ad-bc\\\\0\\\\0\end{bmatrix}$$ 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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so it is quite easy to see the kernel of $F$, you just have to look for $M \in M_2$ so that $det(M)=0$ ... |
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