Linear Algebra If someone could do parts abc that would be a

Linear Algebra

If someone could do parts a,b,c that would be a great help, thanks!

http://postimg.org/image/wfxecdbb5/

Solution

First rotate the plane so that the line is now on the x-axis.

clearly this line has slope b/a, in fact, it is the line y = (b/a)x

(unless a = 0, in which case the line is the y-axis). we can write the

matrix for this rotation as R =

[ a/(a^2+b^2) b/(a^2+b^2)]
[-b/(a^2+b^2) a/(a^2+b^2)].

now to reflect across the line (which is now the x-axis)
we just apply the matrix S =

[1 0 ]
[0 -1]

which sends (x,y) to (x,-y).

to \"rotate back\", we can apply R^-1:

[a/(a^2+b^2) -b/(a^2+b^2)]
[ b/(a^2+b^2) a/(a^2+b^2)].

so the reflection is just R^-1SR, which, being a matrix,

is obviously a linear transformation.

we have already found a formula for the matrix, we just need to do the multiplication:

SR =

[ a/(a^2+b^2) b/(a^2+b^2)]
[b/(a^2+b^2) -a/(a^2+b^2)], so R^-1SR =

[(a^2-b^2)/(a^2+b^2) ....2ab/(a^2+b^2)....]
[....2ab/(a^2+b^2).... (b^2-a^2)/(a^2+b^2)].

if we call this matrix A, then we have det(A-xI) =

x^2 - (a^2-b^2)^2/(a^2+b^2)^2 - 4a^2b^2/(a^2+b^2)^2 =

x^2- (1/(a^2+b^2)^2)(a^4 - 2a^2b^2 + b^4 + 4a^2b^2) =

x^2 - (1/(a^2+b^2)^2)(a^4 + 2a^2b^2 + b^4) =

x^2 - 1 = (x + 1)(x - 1), so the eigenvalues of A are 1 and -1.

so what is an eigenvector corresponding to 1? we set A - I = 0, so

[-2b^2/(a^2+b^2) 2ab/(a^2+b^2)][x]....[0]
[2ab/(a^2+b^2) -2a^2/(a^2+b^2)][y] = [0]

yielding the two equations:

ay - bx = 0
bx - ay = 0, both of which give y = (b/a)x, the line of reflection.