I know how to find which maps are linear, but how do I get representation matrix for linear maps? I know $f(x) := Ax$ . Any explanation would be highly appreciated. I'm really confused. Please help me!
For example, look at this problem.
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Sign up to join this communityI know how to find which maps are linear, but how do I get representation matrix for linear maps? I know $f(x) := Ax$ . Any explanation would be highly appreciated. I'm really confused. Please help me!
For example, look at this problem.
Suppose you have a linear map. In general,
$$ \begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}ax + by\\cx + dy\end{pmatrix}. $$
Then you know that that
$$ \begin{pmatrix}e & f \\ g & h\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}= \begin{pmatrix}ex + fy\\gx + hy\end{pmatrix}=\begin{pmatrix}ax + by\\cx + dy\end{pmatrix}. $$
In particular, you have that $e = a$, $f = b$, $g = c$, and $h = d$, thus your matrix is:
$$ A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}. $$
Once you have the matrix, the map is a rotation if the determinant of $A$ is $1$, and it is orientation preserving if the the determinant of $A$ is positive.
In the first instance of your example this yields:
$$ A_1 = \begin{pmatrix}0 & -2 \\ 1 & 0\end{pmatrix}. $$
To verify just multiply:
$$ \begin{pmatrix}0 & -2 \\ 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0\cdot x +(-2)\cdot y \\ 1\cdot x + 0\cdot y\end{pmatrix} = \begin{pmatrix}-2y \\ x\end{pmatrix} $$
as desired.