If the kernel of a linear transformation T, denoted as ker T, is not equal to zero (ker T ≠ 0), then T is definitively not an isomorphism. This conclusion arises from the fact that a non-zero kernel indicates the existence of multiple solutions for which T maps to zero, thereby demonstrating that T is not injective. Consequently, since injectivity is a necessary condition for a transformation to be an isomorphism, T cannot be isomorphic if ker T is non-zero.
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nautolian said:Homework Statement
Show that if ker T != 0 then T is not an isomorphism.
Homework Equations
The Attempt at a Solution
If Ker T != 0 that means that there are multiple solutions for which T=0 meaning it is not injective and hence not isomorphic? Is that correct? I don't think it is, or is there a better way of proving this? Thanks.