Nice little proof, can any one do it?

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A linear transformation F is one-to-one if F(u) = F(v) implies u = v, which is equivalent to stating that the kernel of F contains only the zero vector, Ker(F) = {0}. To prove this, one can use a contradiction approach: assume F is one-to-one and has non-zero vectors in its kernel, leading to the conclusion that F cannot be one-to-one. The proof also involves demonstrating that F(0) = 0, confirming that the zero vector is in the kernel. The discussion emphasizes that the proof relies on basic properties of linear transformations rather than complex algebra. Ultimately, understanding these properties is crucial for establishing the relationship between one-to-one transformations and their kernels.
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A linear transformation F is said to be one-to-one if it satisfies the following condition: if F(u) = F(v) then u = v. Prove that F is one-to-one if and only if Ker(F) = {0}.
 
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Well, what are your thoughts about this?
 
To help you along a bit:

a) IF one-to one..
You are to show that this implies the kernel of F contains just the 0 element.

Now, you can prove that this must be true, by way of contradiction:

ASSUME that the i) linear transformation F is both ii) one-to-one AND has iii) non-zero vectors in its kernel.

Show that i)+iii) implies that F is NOT one-to-one!
 
hmmm yeah my friend posed this proof to me and asked me to try and find a good solution to it if i could, to be honest i have idea! its possible that i just lack the algebra to do so
 
Um, this proof doesn't really require any algebra, just the basic properties that define a linear transformation. For the reverse direction, suppose F(u) = F(v) so that F(u) - F(v) = F(u-v).

For the forward direction, you could also prove that F(0) = 0, which means that {0} is contained in Ker(F), so you just have to prove that Ker(F) is contained in {0}. Let v be in Ker(F), find out what this means and remember that you should have proved 0 = F(0).
 
ah thanks snipez90, pretty sure I've got it out!
 
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