Question

Help. Image below. Please explain. I am trying to understand Linear Algebra.

Answer 1

Suppose \(T:V\rightarrow W\) is a linear map where \(\dim(V)=n\). Suppose also that \(\text{Rank}(T)=n\). By the Rank-nulity theorem, we have \(\text{Rank}(T)+\text{null}(T)=n\) and thus \(\text{null}(T)=0\). This shows that the kernel contains only the zero vector. We also have \(\ker(T)=\{0\}\) if and only if \(T\) is injective. The result follows.

Answer 2

I am still confuse

Answer 3

T is injective if the only vector maped onto 0 in W is the 0 in V, and this is true in your quetion becuase the rank of your transformation T is the same as your dimension of V

Answer 4

yes but how about the rank? what if it had a different dimension? will it still be true?

Answer 5

dimension was arbitrary