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Show that the mapping f: R2->R2 given by f(x,y)=(x+y,x-y) is linear. For each subspace X of R2 describe f-->(X) and f<--(X).
My question is whether f-->(X) means that the image of the x-axis is the line y=x and the image of the y-axis is the line y=-x, so how am I describe f<--(X)?
Thanks in advance.
Quote:
Originally Posted by jojo7777777
The image of the x-axis (points of the form (x, 0)) is the set (x+ 0, x- 0)= (x, x) so y= x, and the image of the y-axis (points of the form (0, y)) is (0+ y, 0- y)= (y, -y) so y= -x.Quote:
Originally Posted by jojo7777777
The inverse image of a set X is the set of all points (x, y) such that f(x,y) is in X. If f(x,y)= (x+ y, x- y)= (u, v) then x+ y= u, x- y= v. Adding the two equations 2x= u+ v so x= (u+ v)/2. Subtracting x- y= v from x+ y= u, 2y= u- v so y= (u- v)/2. That is,which is the same as
.
since we are dealing with R2, which is finite-dimensional, we can use matrices. of course, to do that, we need to pick a basis, so we can use that basis to turn f into a matrix.
but the standard basis B = {(1,0),(0,1)} will do just fine.
note that f(1,0) = (1,1), and f(0,1) = (1,-1), so relative to the standard basis f has the matrix:
.
this has determinant -2 ≠ 0, so it has an inverse matrix M-1, which corresponds to f-1 (relative to the same (standard) basis).
the inverse is found to be:
.
thus f-1(x,y) = (1/2)(x+y,x-y), as the previous posts have already indicated.
R2 has a particularly simple subspace structure:
if U is a subspace of R2, there are 3 possibilities:
U = R2 (the only 2-dimensional subspace)
U = {a(x0,y0): a in R}, the set of all scalar multiples of a particular non-zero vector (x0,y0) of R2 (1-dimensional subspaces)
U = {(0,0)} (the only trivial subspace).
since f is bijective, if U = R2, then f(U) = f(R2) = R2, and f-1(U) = f-1(R2) = R2.
similarly if U = {(0,0)}, f(U) = {(0,0)} and f-1(U) = {(0,0)}.
so the only "interesting" case is where U = {a(x0,y0): a in R}.
by the linearity of f (or by the linearity of the matrix for f, which is the same thing), f(U) = {a(x0+y0,x0-y0): a in R}.
for example if (x0,y0) = (3,4), f(U) has basis {(7,-1)}.
interestingly enough, f-1 = (1/2)f, so f-1(U) = f(U) (since for any non-zero vector v, {v} and {(1/2)v} generate the same subspace).
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while the x-axis and the y-axis are indeed each subspaces of R2 (as is any line through the origin), they are by no means the ONLY subspaces of R2.
Thank you very much...your answers are explained clearly and help to put in order the information I have...
Just a point I wish to make sense out of it...According to the book (whence this question) the answer involves "the image of the line y=mx where m ≠ 1 is the line y=x(1+m)/(1-m)", how does it fit?
claim: the subset U = {(x,y) in R2: y = mx} is a subspace of R2 of dimension 1.
proof:
we must show 3 things:
a) if (u,v) and (u',v') are in U, so is (u,v) + (u',v').
b) if c is any real number, and (u,v) is in U, so is c(u,v).
c) (0,0) is in U.
we start with (a). if (u,v) and (u',v') are in U, then v = mu, and v' = mu'.
thus (u,v) + (u'v') = (u,mu) + (u',mu') = (u+u',mu+mu'), and mu+mu' = m(u+u'), so (u,v) + (u'v') is in U.
now (b): if (u,v) is in U, then v = mu. so c(u,v) = c(u,mu) = (cu,c(mu)). since c(mu) = (cm)u = (mc)u = m(cu), we see c(u,v) is also in U.
and finally, (c): (0,0) = (0,m0), so (0,0) is in U.
this shows U is indeed a subspace of R2. to prove it has dimension 1, we need to find a basis.
i claim {(1,m)} is a basis for U. clearly this is linearly independent, since (1,m) is a non-zero vector. so all we have to do is show {(1,m)} spans U.
again, suppose that (x,y) is any point of U. then, by the definition of U, y = mx, so (x,y) = (x,mx). thus (x,y) = (x,mx) = x(m,1), so {(m,1)} spans U.
claim 2: if U is a 1-dimensional subspace of R2, then either U is the line y = mx, for some real number m, or U is the y-axis.
suppose U is not the y-axis. since U is one dimensional, some non-zero vector (a,b) is a basis for U. suppose that a ≠ 0.
let m = b/a. then for any vector (x,y) in U, (x,y) = c(a,b) = (ca,cb). thus y = cb = c(b/a)a = (b/a)(ca) = m(ca) = mx, that is: U is the line y = mx.
on the other hand, suppose our basis for U is (0,b), where b ≠ 0. then U = {(x,y) in R2: (x,y) = c(0,b), for some c in R}.
but c(0,b) = (0,cb), so every point of U lies on the y-axis, and every point of the y-axis is in U, so U = the y-axis.
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so yes, the 1-dimensional subspaces of R2 are "lines through the origin" (perhaps explaining why its called "linear" algebra: bases consisting of a single vector are LINES, lines are the "building blocks" by which we make "bigger spaces" (planes, and 3-spaces, etc.)).
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now a linear map, takes lines (through the origin) to "some other lines" (also through the origin).
consider, what happens when we take f and apply it to the vector (x,mx) (that is: a point lying on the line y = mx).
we get f(x,mx) = (x + mx,x - mx).
which line is this?
we can use the "two-point" formula for a line:
which two points shall we use?
how about (x1,y1) = f(0,0) = f(0,m0) = (0+m0,0-m0) = (0,0), and:
(x2,y2) = f(1,m) = (1+m,1-m)
then our line formula becomes:
which simplifies to y = [(1-m)/(1+m)]x (perhaps there is a typo in your book?).
the "bad value" for m would be m = -1:
in this case, f maps (x,-x) to (0,2x), which lies on the y-axis (a line of "infinite slope", that is: vertical).
I am grateful to you for your help...! It is a perfect explanation!