Respuesta :
Answer:
It is proved that there exist a basis of V and a basis of W such that with respect to these bases, all entries of M(T) are 0 except that the entries in row j , column j , equal 1 for 1 <= j <= dimrange T.
Step-by-step explanation:
Given V and W are finite dimentional vector space such that [tex]T\in L(V,W)[/tex] where T is the corresponding lineat transformation from V to W.
To prove the requirment let [tex]u_1, u_2,.......,u_m[/tex] be a basis of dim Ker(T).
Now extend this basis of V to n such that, new basis is of the form [tex]u_1,u_2,.........,u_m, v_1,v_2,..............................,v_n[/tex]. Then basis of range T is of the form, [tex]Tu_1,T_2,.............................,Tu_m,Tv_1,...............,Tv_n [/tex] which are linearly independent. Therefore according to rank-nullity theorem dim range T=n.
Now extending basis of range space into :
[tex]Tv_1,Tv_2,.......,Tv_n,Tu_1,........,Tu_m, w_1,w_2,....................,w_q[/tex] with respect to the basis [tex]v_1,v_2,......,v_n,u_1,u_2,......,u_m[/tex] (note the reverse order of vectors) of V, the matrix of T has the desired form.
Since [tex]Tv_i=1\times Tv_i[/tex] for any [tex]i\in \{1,2,3,4,.....,n\}[/tex] we have all the entries in the first n-column 0, except the entries in row i, column i, equal to 1 for [tex]1\leq i \leq n[/tex]=dim range (T).
[tex]Tu_j=0[/tex] for any [tex]j\in \{1,2,3,......,m\}[/tex]
Therefore all the entries in rest of m columns are 0.
