I've just started Linear Algebra
I have one question i'd like some help with.
The line l is described by the equation
Find an equation for the line in
suppose we knew how to transform e'-coordinates into e-coordinates, by using some matrix A. it stands to reason that A^-1 would translate e-coordinates to e'-coordinates.
now the matrix that sends e'-coordinates to e-coordinates is very easy to write down. in e'-coordinates, e1' is (1,0). and for ANY 2x2 matrix A, A(1,0) is the first column of the matrix. similarly, in e'-coordinates, e2' is (0,1), and A(0,1) is the second column of the matrix A. so the first column of A is A(e1') = (5/13, 12/13) (in e-coordinates). and the second column of A is e2' in e-coordinates, or (-12/13,5/13) so A =
by a standard computation, A^-1 =
[ 5/13 12/13]
so in e' coordinates, e1 = (5/13,-12/13) = (5/13)e1' - (12/13)e2', while e2 = (12/13,5/13) = (12/13)e1' + (5/13)e2'.
in e'-coordinates the points (x1,x2) that lie on the line now have coordinates ((5/13)x1+(12/13)x2, (5/13)x2 - (12/13)x1)), so the equation for our line becomes:
2(5/13)x1 + 2(12/13)x2 + 3(5/13)x2 - 3(12/13)x1 = 0, or upon simplfying:
-2x1 + 3x2 = 0
is the same as saying . That is, that the line is perpendicular to the vector
Given that and , then the matrix having those coefficients as columns,
will change a vector in and components to the same vector in and components. What we want to do is change the vector to and components. We can do that by using the inverse matrix which happens to be
So we are saying that our vector, in components must be perpendicular to
Now, multiply both sides of that equation by 13.
Essentially the same thing but slightly different point of view:
solve , for and in terms of and . (That is equivalent to finding the inverse matrix above.) Once you have done that, replace and in by their expression in terms of and to find the vector our must be perpendicular to.