# Example - linear map

We have the values of a linear map from a vector space $L_{1}$ to a vector space $L_{2}$ on basis B of the vector space $L_{1}$. One of these 5 statements is true, all the others are false. Which one is it?

a) It is possible to calculate a kernel of the linear map but outside of kernel we don't know the values of the linear map.

b) Values of the linear map are determined for the whole domain of definition of $L_{1}$.

c) We don't have enough information to calculate the value of the linear map of any point of $L_{1}$.

d) Values of the linear map on the whole $L_{1}$ are determined only if the linear map is an isomorphism.

e) If the linear map is not an injection, values of the linear map on the whole $L_{1}$ are not determined.

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Linear map or linear transformation? If $T$ be a linear transformation and $B$ is a basis for $L_1$ then clearly, $T(B)$ is a PART of a basis for second vector space $L_2$. – Babak S. Dec 28 '12 at 8:30
From Wikipedia: In mathematics, a linear map, linear mapping, linear transformation, or linear operator (in some contexts also called linear function) is a function between two modules (including vector spaces) that preserves the operations of module (or vector) addition and scalar multiplication............ In this case, it is a linear map from one vector space to another vector space. – user1111261 Dec 28 '12 at 8:46
@BabakSorouh: I dont think so,T(B) is a always a part of basis for $L_2$.What about zero linear transformation? – Shraddha Srivastava Dec 28 '12 at 9:40

HINT: Every $v\in V$ can be written in the form $v=\alpha_1b_1+\alpha_2b_2+\ldots+\alpha_nb_n$ for some scalars $\alpha_1,\alpha_2,\dots,\alpha_n$, where $b_1,\dots,b_n\in B$. If the linear map is $T$, we have
$$T(v)=\alpha_1T(b_1)+\alpha_2T(b_2)+\ldots+\alpha_nT(b_n)\;.$$
We know $T(b)$ for every $b\in B$. Can we determine $T(v)$ for every $v\in V$?
So c is not correct because of your way?? And d is not correct because we don't need $T$ to be isomorphism?? – Babak S. Dec 28 '12 at 9:32
It should be $b$. – Shraddha Srivastava Dec 28 '12 at 9:35