Linear Transformation questions about dimensions

potmobius
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1. Say you have a linear transform from A to B, and where A has a higher dimension than B. How do you show that the kernel of the transform has more than one element (i.e. 0)? Also, if B has a higher dimension than A, then how to show that the transform isn't surjective?

2. The attempt at a solution

By showing that the kernel has more than the element 0, I want to show that the transform isn't injective. But I'm not quite sure how to get there just by using the fact that A has a higher dimension than B. Is that a good way(as in, not too complicated) of proving it? Any ideas?

For the other part, it makes sense intuitively, since the basis of A will have less elements than the basis of B, so there shouldn't be a surjection. But how do you proceed from there to show that the image of A is a proper subset of B?
 
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Think of the theorem that says the dim(A)=dim(Ker(f)) + dim(Im(f)), where f:A->B is a linear morphism (transformation)
 
Thanks! Now I feel really stupid for not considering Rank-Nullity before asking this...
 
potmobius said:
Thanks! Now I feel really stupid for not considering Rank-Nullity before asking this...

You're welcome. I feel stupid for not knowing that theorem has a name ...
 
Do you know any bad math jokes related to this?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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