Find the Matrix A for the linear mapping:
First by reflecting on the plane $5x-1y+5z=0$, then by projecting on the plane $2x + y + z=0$.
So my thought to do this problem is to find the Matrices for both and then taking the produkt of them.
$$B=$$ Reflection Matrix. $$C=$$ Projection Matrix.
To find the reflection:
$(x,y,z) =$ Some vector from $\mathbb R^3$.
I use that the reflection = $(x,y,z) - 2$(projection of $v$ on the normal of the plane)
Where $v$ is a vector and the normal is $(5,-1,5)$.
I get $(1/11)(-39x + 11y -50z, 10x + 9y +10z, -50x +10y-39z)$.
For the projection we see that the origin is in the plane by looking at the equation for the plane. Then we should get the projection by:
Projection of $u$ on the plane $= u -$ projection of $u$ on the normal of that plane.
i get : $(1/6)(2x-2y-2z,-2x+5y+z,-2x-y+5z)$
So then Matrix $A$ should be the same as $CB$. Although I get this to be wrong.
Can someone show me how to do it?
I think your method will work, as soon as you clean up mistakes in the computation of $B$ and $C$...
For instance, for $B$, take vector $(x,y,z)$, and project onto the normal direction: $$\frac1{\sqrt{51}}(5x-y+5z)\frac1{\sqrt{51}}(5,-1,5)$$.
Then subtract twice the result from $(x,y,z)$: $$(x,y,z)-2\frac1{51}(5x-y+5z)(5,-1,5)=-\frac1{51}(-x-2y-50z,10x-49y+10z,-50x-2y-z)$$.
Then write as a matrix, presumably...
Get: $$-\frac1{51}\begin{pmatrix}-1&-2&-50\\10&-49&10\\-50&-2&-1\end{pmatrix}$$.
For C), we get $$\frac1{\sqrt6}(2x+y+z)\frac1{\sqrt6}(2,1,1)$$ for the projection onto the normal. Now subtract from $(x,y,z)$: $$\frac16(2x-2y-2z,-2x+5y-z,-2x-y+5z)$$.
So $$C=\frac16\begin{pmatrix}2&-2&-2\\-2&5&-1\\-2&-1&5\end{pmatrix}$$.
Now $$A=CB=\frac1{306}\begin{pmatrix}78&98&-118\\102&-239&151\\-258&43&85\end{pmatrix}$$.