Linear algebra, when does the implications hold?

[bobby2k]

Hi, I have 4 implications I am interested in, I think I know the answer to the first 2, but the last two is not something I know, however they are related to the first 2 so I will include all to be sure.

Assume that T is a linear transformation from from vectorspace A to B.

T: A -> B
A* is n vectors in A, that is A* = {a1, a2, an}

1.
T(A*) linearly independent -> A* linearly independent
If T(A*) is linearly independent, then A* must be linearly independent, without any requirements for T?

2.
T(A*) lindearly dependent -> A* linearly dependent
If T(A*) is linearly dependent, then we can only conclude that A* is linearly independent only if T is 1-1, if T is not 1-1 we can not conlude anything?

3.
span(T(A*))=B -> span(A*)= A
If span(T(A*)) = B, what requirement must we have to conlude that span(A*) = A. That T is 1-1, surejective, both or none?

4.
span(T(A*)) ≠ B -> span(A*) ≠ A*
If span(T(A*)) is not B what must we have to conlude that span(A*) is not A? Will this implication hold if T is 1-1, surjective, both or none?

thanks

 

[johnqwertyful]

1. If c1a1+...+cnan=0, then T(c1a1+...+cnan)=0. So no restrictions.
2. Injectivity is not needed. Consider T: R^3->R^2. T(x,y,z)=(x,y). Surjectivity is not needed. Consider
T: R^2->R^3. T(x,y)=(x,y,0).
3. Let B={0} and you'll see surjectivity isn't enough. 1-1 is. Think of any x in A and consider T(x)=c1T(a1)+c2T(a2)+...+cnT(an).
4. Your last statement is equivalent to span(A*)=A -> span(T(A*)) = B. If dim(A)<dim(B) and you have n=dim(A) vectors, this is not true. So injectivity is not enough. Surjectivity is though. Consider y in B. Then there is an x in A such that T(x)=y. Or T(c1a1+...cnan)=y.

 

[SteamKing]

The word is 'linearly'.

 

[bobby2k]

2. Injectivity is not needed. Consider T: R^3->R^2. T(x,y,z)=(x,y). Surjectivity is not needed. Consider
T: R^2->R^3. T(x,y)=(x,y,0).

Hey, thanks man.
I just have some more questions, are you allowed to prove a general statement with examples?

Also, i am not sure about the example.
T(x,y,z) = (x,y)
if A* = {(1,1,0),(1,1,1)}
then T(A*) = {(1,1),(1,1)}
so T(A*) is linearly dependent, but A* is not, so the implication is false here.

 

[blue_raver22]

The implications in linear algebra hold in certain situations, depending on the properties of the linear transformation T and the vectors in A.

1. The first implication states that if T(A*) is linearly independent, then A* must also be linearly independent. This holds true regardless of any requirements for T. The reason for this is that if T(A*) is linearly independent, then none of the vectors in A* can be written as a linear combination of the others. This means that even when T is not 1-1, the vectors in A* are still linearly independent.

2. The second implication states that if T(A*) is linearly dependent, then A* may or may not be linearly dependent, depending on whether T is 1-1 or not. If T is 1-1, then A* must also be linearly dependent, since the linear dependence of T(A*) implies that there exists a non-trivial linear combination of the vectors in A* that equals the zero vector. However, if T is not 1-1, then A* may still be linearly independent, as there may be a non-trivial linear combination of the vectors in A* that equals the zero vector, but this does not necessarily mean that the vectors themselves are linearly dependent.

3. The third implication states that if the span of T(A*) is equal to B, then the span of A* must be equal to A. In order for this implication to hold, T must be both 1-1 and surjective. This means that for every vector b in B, there exists a unique vector a in A such that T(a) = b. This also implies that the span of T(A*) must be the entire space B, and therefore the span of A* must be the entire space A.

4. The fourth implication states that if the span of T(A*) is not equal to B, then the span of A* may or may not be equal to A. This depends on whether T is 1-1, surjective, both, or neither. If T is 1-1 and surjective, then the span of T(A*) must be the entire space B, which means that the span of A* must also be the entire space A. However, if T is not 1-1 or surjective, then the span of T(A*) may not equal B, and