# Thread: Prove ST=1 iff TS=1

1. ## Prove ST=1 iff TS=1

Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=1 if and only if TS=1.

Could someone get me started on this please? Thanks for any help.

2. Originally Posted by GreenDay14

Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=1 if and only if TS=1.

Could someone get me started on this please? Thanks for any help.

I think what you need is: if V is a fin. dimensional v.s. then a linear transformation is 1-1 iff it is onto, and thus:
A lin. transf. in a fin. dim. v.s. is invertible iff it is 1-1.
Finally, for any two functions $f,\,\,g$: $fg=id. \Longrightarrow g \,\,is\,\,1-1$

Tonio

3. Originally Posted by tonio
I think what you need is: if V is a fin. dimensional v.s. then a linear transformation is 1-1 iff it is onto, and thus:
A lin. transf. in a fin. dim. v.s. is invertible iff it is 1-1.
Finally, for any two functions $f,\,\,g$: $fg=id. \Longrightarrow g \,\,is\,\,1-1$

Tonio
Thank you very much Tonio, if i may ask tho, what do you mean by 1-1?

4. Originally Posted by GreenDay14
Thank you very much Tonio, if i may ask tho, what do you mean by 1-1?

One to one, injective: $f(x)\,\,is\,\,injective\,\,iff\,\,f(x)=f(y) \Longrightarrow x=y$

Tonio

5. Originally Posted by GreenDay14

Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=1 if and only if TS=1.
Sorry, this should read:

Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=I if and only if TS=I.

Notice that the 1's were changed to the letter I. Sorry for any mix up.

6. Originally Posted by GreenDay14
Sorry, this should read:

Suppose that V is finite dimensional and S,T belong to L(V). Prove that ST=I if and only if TS=I.

Notice that the 1's were changed to the letter I. Sorry for any mix up.

It never minds: anyone understanding this stuff won't get confused and, in fact, many times 1 is used in place of I.

Tonio

7. Okay, so I went and tackled this question a different way and came up with:

Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional S must be invertible since it is injective. Thus if ST=I then TS=I.

Assume TS=I. Let v' be an element of V such that Sv' is in the null space of T. v'=Iv'=TSv'=0. Since V is finite dimensional T must be invertible since it is injective. Thus if TS=I then ST=I.

Therefore ST=I if and only if TS=I.

Does this make sense?

8. Originally Posted by GreenDay14
Okay, so I went and tackled this question a different way and came up with:

Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional S must be invertible since it is injective. Thus if ST=I then TS=I.

Assume TS=I. Let v' be an element of V such that Sv' is in the null space of T. v'=Iv'=TSv'=0. Since V is finite dimensional T must be invertible since it is injective. Thus if TS=I then ST=I.

Therefore ST=I if and only if TS=I.

Does this make sense?

You didn't show S is injective: you only showed that if Tv is in Null(S) then v = 0. You should show that v in Null(S) ==> v = 0.

Tonio

9. Originally Posted by tonio
You didn't show S is injective: you only showed that if Tv is in Null(S) then v = 0. You should show that v in Null(S) ==> v = 0.

Tonio
So are you saying that in my first line i am actually showing that t is injective, rather than S? or is neither shown. Also does this mean that my second line is garbage?

10. So could itb e changed to:

Assume ST=I. Let v be an element of V, such that Tv is an element of null S. v=Iv=STv=S(Tv)=0. Since V is finite dimensional T must be invertible since it is injective. Thus if ST=I then TS=I.

and also best to add:

ST = I
TSTT' = TIT' (where T' is the inverse of T)
TS = I

or am i still on the wrong track?

11. This just follows from the fact that composition of invertible linear transformations is associative (function composition in general is associative). If $ab=1$ then

$(ab)a=1a=a$

but by associativity

$(ab)a=a(ba)$

So $a(ba)=a$. Multiplying on the left by $a^{-1}$ on both sides yields $ba=1$.

The above obviously holds for any group.

12. Originally Posted by GreenDay14
So are you saying that in my first line i am actually showing that t is injective, rather than S? or is neither shown. Also does this mean that my second line is garbage?

You're showing neither, the same as in your second line. You could try to prove the following easy lemma from basic set theory:

$ST=I \Longrightarrow T\,\,is\,\,\,1-1\,\,\,and\,\,S\,\,is\,\,\,onto$

Tonio

13. Originally Posted by tonio
You're showing neither, the same as in your second line. You could try to prove the following easy lemma from basic set theory:

$ST=I \Longrightarrow T\,\,is\,\,\,1-1\,\,\,and\,\,S\,\,is\,\,\,onto$

Tonio
I apologize Tonio, but I am unsure as to what you mean by onto? onto what?