help with proofs........?

Question

if dim(V) < dim(W), prove that T cannot be onto (surjective)
and
if dim(V) > dim(W), prove that T cannot be one to one (injective)

Answer 1

~ you need to say what are V and W??
I am guessing that V and W are vector spaces and that T is LINEAR transformation from V --> W.

suppose dim(V) < dim(W), take a basis of W: w1, ... ,wn, and assume that T is onto, then for every wi there exist a vi such that
T(v1)=w1, ... , T(vn)=wn,
now let us check if v1,... vn are linearly independent:
take a1v1 + ... + anvn =0 a linear combination equal to cero, then
T ( a1v1 + ... + anvn ) = T(0) =0,
on the other hand,
T ( a1v1 + ... + anvn ) = a1T(v1) + ...+ anT(vn)
= a1w1 + ... + anwn =0, since w1,..., wn are a basis, that means that
a1=a2=..=an=0
but this implies that v1, ... vn are linearly independent.
and this is a contradiction to the fact that dim V < dim W


for the other problem:
if dim(V) > dim(W), prove that T cannot be one to one (injective).

Assume that v1,..., vm form a basis for V,
if Ker(T) =0 (i.e. if we assume that T is 1-1), then
T(v1), ... , T(vm) are a linearly independent set of vectors, but
m=dim V > dim (W), but this cannot happen, since any set of linearly independent vectors in W would have to have fewer than
dim(W) vectors.

Answer 2

You don't say whether T maps V to W or W to V, but the question makes sense the first way.

Since T is a function (it is, isn't it), dim( T(V)) = dim(V), which just says T produces just 1 value in W for each value in V, not necessarily distinct, which means there will be values in W LEFT OVER such that there is no v in V for which T(v) = w (in W). This is obvious from the definition of function. Remember the vertical line test in high school math? The hard part is wording it precisely in the mathematical language you are learning. Study the examples you've seen and do your best to imitate the style.

Second half of the question is similar. If dim(V) > dim(W), then even though all of V can be mapped onto W, some elements of V will have to be mapped onto the same elements of W, so T inverse won't know where to map them back. Remember the horizontal line test from high school? Bottom line, to have an inverse, a function must be both 1 to 1 and onto.

Answer 3

1) With T:V -> W as a linear function, assume dim(V)=n and dim(W)=m. Since every vector in V can be written as a linear combination of n vectors, and since T is linear, we can show that any vector in the image of T can be written as a combination of n vectors in W. Since the dimension of W is greater than the dimension of V, the image of T can't cover W.

Say {v1, v2, ..., vn} is a basis of V. Let y be in the image of T. Then there are some c1, ..., cn so that

y = T(c1 v1 + ... + cn vn) = c1 T(v1) + ... + cn T(vn).

Thus any vector in the image of T is a linear combination of T(v1), ..., T(vn). Since n < m, this set does not form a basis of W, therefore T can't be surjective.

2) Again let dim(V)=n and dim(W)=m, with {v1, ... , vn} a basis of V, but now let n > m. Since there are more than m items in the list T(v1), T(v2), ... , T(vn), this collection of vectors is linearly dependent in W, so there exist constants c1, c2, ..., cn so that c1 T(v1) + c2 T(v2) + ... + cn T(vn) = 0 where at least one of the c's is not zero. By linearity of T we can rewrite this as

T(c1 v1 + c2 v2 + ... + cn vn) = 0.

Since at least one of the c's is nonzero and the v's form a basis of V, we conclude that the sum c1 v1 + c2 v2 + ... + cn vn is nonzero. Thus there is a nonzero vector of V which T sends to zero, so the kernel of T is nontrivial. Thus T is not 1-1. Cheers!