# Linear Transformations

1. Nov 20, 2005

### playboy

Let T: V-->W be a linear transformation.

a) If T is one-to-one and TR=TR1 for transformations R and R1: U -->V, show that R = R1
b) If T is onto and ST=S1T for transformations S and S1: W -->U, show that S=S1

I am sooo very lost here, and no idea where to start:(

for part a) what does it mean by TR and TR1?
just T(R): U --->V and T(R1): U ---V?

If its onto, dosn't that just mean T(R) = o and then show that R = 0?

I wont even bother with part be given that im confused with part A

Thanks

2. Nov 20, 2005

### AKG

Definitions:

If A is a map from X to Y, and B is a map from Y to Z, then BA is a map from X to Z defined by:

BA(x) = B(A(x))

for all x in X. So, if f(x) = 3x, and g(x) = x+5, then gf(x) = g(f(x)) = g(3x) = 3x+5. Note that fg(x) = f(g(x)) = f(x+5) = 3(x+5) = 3x+15, so gf is not the same as fg.

If A is a map from X to Y and B is also a map from X to Y, then A = B if and only if, for all x in X, A(x) = B(x).

A map A from X to Y is one-to-one if and only if, for all x and x' in X, A(x) = A(x') if and only if x = x'. In simpler terms, a one-to-one map never sends two different things to the same element; each element of the co-domain gets its own image. The map f on the reals defined by f(x) = x² is not one-to-one, because it sends both 1 and -1 to the same thing, namely 1.

A map A from X to Y is onto if and only if, for all y in Y, there exists some x in X such that A(x) = y. In simpler terms, a map is onto if it reaches everything in the codomain. The map f in the previous example is not onto, since it will only map into the non-negative half of the reals.

3. Nov 22, 2005

### playboy

no progress so far :(

4. Nov 23, 2005

### HallsofIvy

Staff Emeritus
No, R takes a vector in U to one in V, then T takes it to a vector in W.
TR: U-> W That is, if x is in U, apply R to it to get R(x) in V. Now apply T to that to get T(R(x)) in W. It is essentially composition of functions.
No, "one-to-one" means that if T(x)= T(y) then x= y: two different vectors in V cannot[\b] be mapped to the same vector in W. For T a linear transformation, T(0)= 0 so "one-to-one" for a linear transformation is the same as saying T(x)= 0 if and only if x= 0.
TR= TR1 means that for every vector x in U, TR(x)= TR1(x). That is,
T(R(x))= T(R1(x)). Since T is one-to-one, what does that tell you about
R(x) and R1(x)? Since x could be any vector in U, what does that tell you about R and R1?

"onto" means that the range of T is all of W: if y is any vector in W, then there is some x in V so that T(x)= y. (for example, x2 is not "onto" the set of all real numbers since no x2 is equal to -1. It is "onto" the set of all non-negative numbers.)

Again ST= S1T1 means that S(T(x))= S1(T(x)) for all x in V.
Let y be any vector in W. Then there exist x such that y= T(x). For that x, S(T(x))= S(y)= S1(T(x))= S1(y). That is, for every y in W, S(y)= S1(y).

5. Nov 23, 2005

### playboy

Since T is one-to-one, that tells me that R(x) = R1(x)
And since X is any vector in U, then that tells me that R = R1

Last edited by a moderator: Nov 23, 2005
6. Nov 26, 2005

### playboy

Okay im back again with a similar question.

Let V---(T)--->U---(S)---W be linear transformations.
a) If ST is one-to-one, show that T is one-to-one and that dimV </= dim U

Here is how i attempted it:

Assume T is NOT one-to-one

if x is a vector in V, then for ever x not equal to 0 such that T(x) = 0

then
ST(x) = 0
=S(T(x))
=S(0)
therefore, T is one-to-one?

How does that sound?

Now how do i show that dimV is </= dim U?

7. Nov 26, 2005

### HallsofIvy

Staff Emeritus
I'm not sure what you mean by this: "if x is a vector in V, then for ever x not equal to 0 such that T(x) = 0"
It doesn't have to be for "ever(y) x not equal to 0". It is true that if T is not one-to-one, then there must exist some x, not equal to 0, such that T(x)= 0. Then, of course, ST(x)= S(0)= 0 which contradicts the fact that S is one-to-one.

Now, apply T to every basis vector of V. How many vectors does give you? What does that tell you about the dimension of T(V) (which is a subspace of U)?

8. Nov 27, 2005

### playboy

Okay im so lost. Ill start again.

Let x be a basis vector in V, not equal to 0, such that T(x) = 0
Then ST(x) = S(0) = 0 contradicts that S is one-to-one?
How does this show that T is one-to-one

If T is one-to-one, ker(T) = 0. therefore, dimV = dimImage(T) and Image(T) = U... so dimV = dimU

How does that sound?

9. Nov 28, 2005

### playboy

oh i don't know how i missed that, i see why it contradicts one-to-one! im so silly.

Now for the dimension,

If you apply T to ever basis in V, since T is one-to-one, you will get that many vectors. And since V is a subspace of U, its dimension cannot be greater than that of U.

How does that sound? (If anyone actually seems to care :S)