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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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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.

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  • $\begingroup$ Could you explain this taking the first instance of my example? Because I still have some doubts about the second line(dot product). $\endgroup$
    – Isuru
    Jan 28, 2016 at 3:59
  • $\begingroup$ I just added a full walkthrough of your example. $\endgroup$ Jan 28, 2016 at 4:11
  • $\begingroup$ Thanks a lot. I get it now :) $\endgroup$
    – Isuru
    Jan 28, 2016 at 4:20

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