# direct image inverse image

• November 4th 2012, 08:50 AM
jojo7777777
direct image inverse image
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)?
• November 4th 2012, 09:23 AM
Plato
Re: direct image inverse image
Quote:

Originally Posted by jojo7777777
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)?

$\overleftarrow f \left\{ {\left( {x,y} \right)} \right\} = \left( {\frac{{x + y}}{2},\frac{{x - y}}{2}} \right)$
• November 4th 2012, 09:59 AM
HallsofIvy
Re: direct image inverse image
Quote:

Originally Posted by jojo7777777
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)?

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.

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, $f^{-1}(u, v)= ((u+ v)/2, (u- v)/2)$ which is the same as $f^{-1}(x, y)= ((x+ y)/2, (x- y)/2)$.
• November 4th 2012, 12:08 PM
Deveno
Re: direct image inverse image
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:

$M = \begin{bmatrix}1&1\\1&-1 \end{bmatrix}$.

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:

$M^{-1} = \begin{bmatrix}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&-\frac{1}{2} \end{bmatrix}$.

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).

*********

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.
• November 5th 2012, 01:35 AM
jojo7777777
Re: direct image inverse image
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?
• November 5th 2012, 07:59 AM
Deveno
Re: direct image inverse image
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.

***********

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.)).

***********

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:

$y - y_1 = \left(\frac{y_2-y_1}{x_2-x_1}\right)(x - x_1)$

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:

$y - 0 = \left(\frac{1-m-0}{1+m-0}\right)(x - 0)$

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).
• November 6th 2012, 06:58 AM
jojo7777777
Re: direct image inverse image
I am grateful to you for your help...! It is a perfect explanation!