Linear Algebra - Finding the matrix for the transformation

yesiammanu
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Homework Statement


Find the matrix for the transformation which first reflects across the main diagnonal, then projects onto the line 2y+√3x=0, and then reflects about the line √3y=2x


Homework Equations


Reflection about the line y=x: T(x,y)=(y,x)
Orthogonal projection on the x-axis T(x,y)=(x,0)
Orthogonal projection on the y-axis T(x,y)=(0,y)

The Attempt at a Solution


Reflection about the line y=x: T(x,y)=(y,x), so the standard matrix for this would be the matrix {(0,1),(1,0)}

However I'm not sure how to deal with equations rather than axis. I assume in the second projection, you can simplify it to y=√3/2 x. Can you then separate these into a scalar operation (√3/2) and orthogonal operation (y=x)? Even so, I wouldn't know how to go further than this since I only know how to do it among the axis
 
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Any vector that lies along the line 2y+ \sqrt{3}x= 0 is mapped to itself while any vector perpendicular to that is mapped to 0. One vector along that line is (-2, \sqrt{3}) and, of course, (\sqrt{3}, 2) is perpendicular to it. So
\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix} -2 \\ \sqrt{3} \end{bmatrix}= \begin{bmatrix} -2 \\ \sqrt{3} \end{bmatrix}
and
\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}\sqrt{3} \\ 2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}
Those give you four equations to solve for a, b, c, and d.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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