Linear transformation problem

Hi! Have some troubles with solving one problem in my assignment, would be great if somebody could help me. The problem is following:

Let *n>0* and let *L*_{n,k }, whear *1<=k<=n* be a subspace spanned by vectors *e*_{j} for *j=1...k*. Show that it exist a linear transformation T:R^{n}=>R^{(n-k)} such as L_{n,k}=*Ker*(T)

Re: Linear transformation problem

You can always find a **basis** for $\displaystyle L_{n,k}$ contained in $\displaystyle \{e_j\}_{j=1}^k$ then extend it to a basis for $\displaystyle R^n$. Define $\displaystyle T(e_i)= 0$ for $\displaystyle e^i\in \{e^j\}_{j=1}^k$, T(v)= 0 for v in the extended basis. T(v) is defined "by linearity" for all other v in $\displaystyle R^n$. That is, since $\displaystyle \{e_1, e_2,..., e_k, f_1, f_2, ..., f_{n-k}\}$, we can write $\displaystyle v= a_1e_1+ a_2e_2+ ...+ a_ke_k+ b_1f_1+ b_2f_2+ ...+ b_{n-k}f_{n-k}$ and then $\displaystyle Tv= a_1T(e_1)+ a_2T(e_2)+ ...+ a_kT(e_k)+ b_1T(f_1)+ b_2T(f_2)+ ...+ b_{n-k}T(f_{n-k})= a_1+ a_2+ ...+ a_k$. Then Tv= 0 if and only if v is in the subspace.

Re: Linear transformation problem

Thanx for answe! Could you please spicify what do you mean by saying that Tv=0, is it a transformation of v? What is v if it can be represented as v=a(1)e(1)+....b(n-k)*f(n-k). And if you said that T(e_j)=0, how can Tv be equal to a_1+a_2+....+a_k if the all supposed to be transformed into null!?