Proving Injectivity and the Identity Map for Finite Dimensional Linear Maps

[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?
thanks in advance.

 

[Mystic998]

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

 

[Hurkyl]

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?
thanks in advance.

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"?)

 

[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
So, your N(T) = {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.

 

[HallsofIvy]

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.

 

[sihag]

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

 

[Mystic998]

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.

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.

 

[HallsofIvy]

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

 

[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

 

[sihag]

the entire proof is there !
scroll up to my reply.

 

[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.

 

[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?

 

[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.