I'm trying to prove that a linear map is injective

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This discussion centers on the injectivity of linear maps, specifically addressing the condition under which a linear map T is injective. It is established that a linear map T is injective if and only if its kernel, denoted as ker(T), contains only the zero vector, {0}. The participants clarify that if T maps two distinct vectors u and v to the same output, then T(u - v) = 0, indicating that u - v is a non-zero vector in the kernel, thus proving T is not injective.

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catcherintherye
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hello, I've been reading some proofs and in keep finding this same argument tyo prove that a linear map is injective viz, we suppose that t(a,c) = 0 and then we deduce that a,c = 0,0. is it the case that the only way a linear map could be non injective is if it took two elements to zero? i.e. t injective iff ker(t) not zero?
 
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First, it's easy to prove that any linear map takes the 0 vector to 0. If a linear maps takes some non-zero vector to 0 also, then it clearly is not injective.

On the other hand, suppose the linear map T takes two vectors, u, v, with u\ne v, into the same thing. Then T(u- v)= T(u)- T(v)= 0 so T takes the non-zero vector u-v into 0.

However, your final statement is exactly reversed: t is injective if and only if ker(t) is {0}.
 

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