# Thread: Linear Algebra- injective, surjective...??

1. ## Linear Algebra- injective, surjective...??

Let U, V, and W be vector spaces over F where F is R or C. Let S: U -> V and T: V -> W be two linear maps. Recall that the composition TS is defined by (TS)(x) = T(S(x)).

(a) Prove that if TS is injective, then S is injective.
(b) Prove that if TS is surjective, then T is surjective.
(c) Assume that TS is bijective. Prove that S is surjective if and only if T is injective.

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(a) Assume that TS is injective. We need to prove that S is injective.
Injective means one-to-one => TS is one-to-one. This implies that TS(x) = 0 where x = 0.
Assume that S(x) = 0. We want to prove that x = 0.
S(x) = 0 => T(S(x)) = T(0) => T(0) = 0 => x = 0. So, S is one-to-one => S is injective.

is this okay?? should i write more?

(b) Assume that TS is surjective. We need to prove that T is surjective.
Surjective means onto => TS is onto W.
Assume R(TS) = W. We want to prove R(T) = W.

I have NO idea how to do this one...

(c) Assume TS is bijective. This means that TS is both injective and surjective => TS is one-to-one correspondence. We need to prove that S is surjective iff T is injective.
I know after proving (a) and (b), we can say that S is injective and T is surjective.

"=>" Assume S is surjective. We need to prove that T is injective.
TS is bijective => TS is injective => S is injective.
But we assumed that S is surjective which then implies that S is an isomorphism.
TS is bijective => TS is surjective => S is surjective.

dim(U) = N(S) + R(S) Because S is an isomorphism, N(S) = 0 and R(S) = dim(V). Thus, dim(U) = dim(V).

dim(U) = N(TS) + R(TS) Because TS is injective, N(TS) = 0 and R(TS) = dim(W). Thus, dim(U) = dim(W).

By Dimension Theorem, dim(V) = dim(W). Thus T is injective.

(Is this right?? can I just conclude that?)

"<=" Assume T is injective. We need to prove that S is surjective.

TS is bijective => TS is surjective => T is surjective. But we assumed that T is injective which implies that T is an isomorphism.
TS is bijective => TS is injective => T is injective.

dim(V) = N(T) + R(T) Because T is an isomorphism, N(T) = 0 and R(T) = dim(W). Thus, dim(V) = dim(W).

Now I'm really confused. I think I just mainly don't understand all this bijective and surjective stuff.

Thanks so much to those who help me with this problem. I really need it.

2. Originally Posted by star637
Let U, V, and W be vector spaces over F where F is R or C. Let S: U -> V and T: V -> W be two linear maps. Recall that the composition TS is defined by (TS)(x) = T(S(x)).
Originally Posted by star637
(a) Prove that if TS is injective, then S is injective.
(b) Prove that if TS is surjective, then T is surjective.
(c) Assume that TS is bijective. Prove that S is surjective if and only if T is injective.
Originally Posted by star637
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(a) Assume that TS is injective. We need to prove that S is injective.
Injective means one-to-one => TS is one-to-one. This implies that TS(x) = 0 where x = 0.
Assume that S(x) = 0. We want to prove that x = 0.
S(x) = 0 => T(S(x)) = T(0) => T(0) = 0 => x = 0. So, S is one-to-one => S is injective.

is this okay?? should i write more?
i don't know.. but why did you choose an arbitrary element whose image when it was applied to the transformation became zero? do you think, the image should also be arbitrary(or so)?

now, suppose TS in injective. if x,y in V such that x=y and there is an a,b such that x=S(a) and y=S(b), then T(x) = T(y).. so, T(S(a)) = (TS)(a) = (TS)(b) = T(S(b)). since TS in injective, then a=b. therefore, S in injective.. QED

Originally Posted by star637
(b) Assume that TS is surjective. We need to prove that T is surjective.
Surjective means onto => TS is onto W.
Assume R(TS) = W. We want to prove R(T) = W.

I have NO idea how to do this one...
Suppose TS is sujective.. if w is in W, then there is a u in U such that (TS)(u)=w. then there is a v in V such that v=S(u). threfore (TS)(u) = T(S(u)) = T(v) =w.. hence, T is surjective..

3. Originally Posted by star637
(c) Assume TS is bijective. This means that TS is both injective and surjective => TS is one-to-one correspondence. We need to prove that S is surjective iff T is injective.
I know after proving (a) and (b), we can say that S is injective and T is surjective.

"=>" Assume S is surjective. We need to prove that T is injective.
TS is bijective => TS is injective => S is injective.
But we assumed that S is surjective which then implies that S is an isomorphism.
TS is bijective => TS is surjective => S is surjective.

dim(U) = N(S) + R(S) Because S is an isomorphism, N(S) = 0 and R(S) = dim(V). Thus, dim(U) = dim(V).

dim(U) = N(TS) + R(TS) Because TS is injective, N(TS) = 0 and R(TS) = dim(W). Thus, dim(U) = dim(W).

By Dimension Theorem, dim(V) = dim(W). Thus T is injective.

(Is this right?? can I just conclude that?)

"<=" Assume T is injective. We need to prove that S is surjective.

TS is bijective => TS is surjective => T is surjective. But we assumed that T is injective which implies that T is an isomorphism.
TS is bijective => TS is injective => T is injective.

dim(V) = N(T) + R(T) Because T is an isomorphism, N(T) = 0 and R(T) = dim(W). Thus, dim(V) = dim(W).

Now I'm really confused. I think I just mainly don't understand all this bijective and surjective stuff.

Thanks so much to those who help me with this problem. I really need it.
i think, that is too long..

Let TS be bijective.
If S is surjective, then S in onto V, that is for every v in V, there is a u in U such that S(u)=v.
since TS is bejective, then S is also injective (from the previous item), that is S is one to one.. so if (TS)(u1) = T(S(u1)) = T(S(u2)) = (TS)(u2) for u1,u2 in U, then u1=u2 (TS is bijective) and so S(u1)=S(u2) (S is injective) and these S(u1) and S(u2) are in V. thus, T is injective..

can you do for the other one? it is just similar..

4. Originally Posted by kalagota
i think, that is too long..

Let TS be bijective.
If S is surjective, then S in onto V, that is for every v in V, there is a u in U such that S(u)=v.
since TS is bejective, then S is also injective (from the previous item), that is S is one to one.. so if (TS)(u1) = T(S(u1)) = T(S(u2)) = (TS)(u2) for u1,u2 in U, then u1=u2 (TS is bijective) and so S(u1)=S(u2) (S is injective) and these S(u1) and S(u2) are in V. thus, T is injective..

can you do for the other one? it is just similar..
well the injective i pretty much understand...it was mainly the surjective one that i actually didn't understand how to do...so to answer your question...no not really.

5. ok.. maybe, this would be my last activity for today.. it is almost 12midnight here.. Ü

so suppose T is injective..

TS is bijective, implies T is surjective (from the previous). so, if w is in W, then there is a v in V such that T(v)=w.
Also, there is a u in U such that (TS)(u) = w (the same w we had for T(v). and so, (TS)(u)=T(S(u)) = w
therefore, since T is injective, we have T(v) = T(S(u)) implies v = S(u).. since v was arbitrarily chosen and we had produce a u in U, then S is surjective.. QED..

6. If ToS is injective, then suppose that S(a)=S(b).
By definition T(S(a))=T(S(b)) or ToS(a)=ToS(b).
But we know that ToS is injective so a=b.
That proves that S is injective.

If ToS is surjective, then if w is in W because ToS is surjective there is some a in U such that ToS(a)=w. But S(a) is in V and T(S(a))=v therefore T is surjective.

Now suppose that ToS is bijective and S is surjective.
If T(c)=T(d) there points a & b is U such that S(a)=c and S(b)=d because S is surjective. This means T(S(a))= T(S(b)) or a=b because of bijectivity or c=d. Which proves T is injective.

On the other hand, suppose ToS is bijective and T is injective.
If p is a point in V, T(p) is a point in W. By bijectivity there is a point k in U such that ToS(k)=T(p). Now we have T(S(k))=T(p) which means that S(k)=p by injectivity. But this shows that S is a surjection.

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### if ToS is bijective map then

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