# Identity map and injectivity

1. Feb 25, 2008

### mind0nmath

The question is to prove for finite dimensional T: V to W,
T is injective iff there exists an S: W to V such that ST is the identity map on V.
I can't quite make the connection between injectivity and the identity map.
any suggestions?

2. Feb 25, 2008

### Mystic998

What's the definition of injectivity? How could you use that to define a function from W to V?

3. Feb 25, 2008

### Hurkyl

Staff Emeritus
Have you tried working with any specific examples to get ideas? Can you think of any examples of injective linear maps?

(I'm assuming V and W are supposed to be vector spaces, and T and S are supposed to be linear transformations)

(What does it mean for T to be "finite dimensional"?)

4. Feb 25, 2008

### sihag

Suppose ST = I
To show T is injective:
Let T(x)=T(y) , f.s. x,y in V
=> S(T(x)) = S(T(y)) [permitted since S is well defined from W to V, and T is of course from V to W]
=> (ST)(x) = (ST)(y)
=> I(x) = I(y)
=> x = y
proving, T is injective.

I think the converse is true only when Dim V = Dim W

Now, suppose T is injective.
Let x be arbitrary in N(T)
T(x) = 0 (zero of W) = T(0)
=> x = 0
Now Dim N(T) + Dim R(T) = Dim V (Sylvester's law)
But N(T) = 0
=> Dim R(T) = Dim V = Dim W
Also R(T) is contained in W
Therefore, R(T) = W
=> T is surjective.

Now T is bijective => T is invertible, so there exists an S from W to V s.t ST = TS = I

Q.E.D.

5. Feb 25, 2008

### HallsofIvy

Staff Emeritus
You are assuming that U and V are finite dimensional vector spaces and T is a linear transformation from U to V?

Saying that there exist S such that ST= I means that T has an inverse. That is not possible unless T is, in fact, bijective: both injective and surjective. It is fairly easy to prove that if a linear transformation is injective, then it is also surjective, and vice-versa.

6. Feb 25, 2008

### sihag

erm, where did my reply go ?
this is strange.

7. Feb 25, 2008

### Mystic998

I don't think this is true unless you assume U and V have the same dimension. For T to be injective, it must be the case that dim V $\geq$ dim U, but equality doesn't have to hold. However, there does have to be an S such that ST is the identity on U (namely you just take Sx to be the preimage under T if x is in the image, and take Sx = 0 otherwise).

Of course, knowing me I'm probably overlooking something painfully obvious, so please be gentle if I'm wrong here.

8. Feb 25, 2008

### HallsofIvy

Staff Emeritus
No, you are completely right. I was thinking of U and V having the same dimension without even realizing it.

9. Feb 27, 2008

### mind0nmath

thanks for all the hints. To clear up: S and T are Linear Transformations. and W is finite dimensional (V and W are vector spaces though nothing mentioned about the dimension of V) . I remember from a logic/proof class that the way to prove if and only if statements is to go both ways meaning:
1) suppose T is injective. Show that there exists a linear transformation S: W -> V such that ST is the identity map on V.
2) Suppose there exists a linear transformation S: W -> V such that ST is the identity map on V. Prove T is injective.
I know that the question is hinting at an inverse of T (namely S), and I know that invertible functions are bijective. so is this proposition false? then what would be the counter example? of if true, how do I go from T is injective to an inverse S???
thanks again

10. Feb 27, 2008

### sihag

the entire proof is there !

11. Feb 27, 2008

### Mystic998

By the way, my example map is completely wrong. That's what happens when you answer algebra questions while doing complex analysis I suppose.

12. Feb 26, 2009

### raizel101

what if V is not finite dimensional? how do you prove that T is injective implies that there exists S such that ST is an identity map?

13. Mar 4, 2009

### ThirstyDog

Hi,

The first half of sihag proof is right but the assumption that:
Dim V = Dim W,
is wrong.

The proof of the other way is:
If T:V->W is injective define
S(x) = y if T(y)=x or,
S(x) = 0 if for all y in V T(y) <> x
Hence
S(T(y)) = S(x) = x => ST is the identity for all element in V.

Just remember that ST = I does not mean TS = I. This is explained as
T: V->W and S:W->V,
ST: V -> V and TS: W -> W
they are different maps.