If Ker T = 0 then T is not isomorphism

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If the kernel of a linear transformation T is not equal to zero, it indicates that there are multiple inputs mapping to the zero vector, which means T is not injective. Consequently, since injectivity is a requirement for a transformation to be an isomorphism, T cannot be an isomorphism. The discussion confirms that if ker T is non-zero, it directly implies T's lack of injectivity and isomorphism. Therefore, the assertion that if ker T ≠ 0 then T is not an isomorphism is validated. This conclusion is essential for understanding the properties of linear transformations in relation to their kernels.
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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.
 
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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.

Well, it is correct. If x is a nonzero element of ker T, then Tx=0=T0. Not injective. Not isomorphism.
 

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