Hi,
D1 and D2 are two vectorial lines of R^2. They are linearly independent. (so they are different lines)
s:R^2->R^2 is the "symmetry of D1 in the direction of D2", im not so sure how to explain it, so let me use a math formula :
if v belongs do D1 and w belongs do D2, and u=v+w then s(u)=v-w
now I have to prove there is a basis in which the matrix of s is :
(line 1) : (1, -1)
(line 2) : (0, -1)
I have absolutely no idea on how to prove it since we don't know a basis of the lines...
I have tried something but it led to nowhere :
v0 belongs to D1, w0 to D2
v0-w0=e1
w0-e1=e2
u=xe1+ye2=x(v0-w0)+y(2w0-v0)=(x-y)v0+(2y-x)w0
so s(u) = (x-y)v0-(2y-x)w0
and then ? v0 and w0 can be anything... two random vectors of R^2... which arent linearly independent, but it doesn't really help
Any idea ? thanks
D1 and D2 are two vectorial lines of R^2. They are linearly independent. (so they are different lines)
s:R^2->R^2 is the "symmetry of D1 in the direction of D2", im not so sure how to explain it, so let me use a math formula :
if v belongs do D1 and w belongs do D2, and u=v+w then s(u)=v-w
now I have to prove there is a basis in which the matrix of s is :
(line 1) : (1, -1)
(line 2) : (0, -1)
I have absolutely no idea on how to prove it since we don't know a basis of the lines...
I have tried something but it led to nowhere :
v0 belongs to D1, w0 to D2
v0-w0=e1
w0-e1=e2
u=xe1+ye2=x(v0-w0)+y(2w0-v0)=(x-y)v0+(2y-x)w0
so s(u) = (x-y)v0-(2y-x)w0
and then ? v0 and w0 can be anything... two random vectors of R^2... which arent linearly independent, but it doesn't really help
Any idea ? thanks