Question

Suppose V and W are finite-dimensional and T ∈ L(V, W). Prove 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.

Answer 1

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 $T\in L(V,W)$ where T is the corresponding lineat transformation from V to W.

To prove the requirment let $u_1, u_2,.......,u_m$ be a basis of dim Ker(T).

Now extend this basis of V to n such that, new basis is of the form $u_1,u_2,.........,u_m, v_1,v_2,..............................,v_n$. Then basis of range T is of the form, $Tu_1,T_2,.............................,Tu_m,Tv_1,...............,Tv_n$ which are linearly independent. Therefore according to rank-nullity theorem dim range T=n.

Now extending basis of range space into :

$Tv_1,Tv_2,.......,Tv_n,Tu_1,........,Tu_m, w_1,w_2,....................,w_q$ with respect to the basis $v_1,v_2,......,v_n,u_1,u_2,......,u_m$ (note the reverse order of vectors) of V, the matrix of T has the desired form.

Since $Tv_i=1\times Tv_i$ for any $i\in \{1,2,3,4,.....,n\}$ we have all the entries in the first n-column 0, except the entries in row i, column i, equal to 1 for $1\leq i \leq n$=dim range (T).

$Tu_j=0$ for any $j\in \{1,2,3,......,m\}$

Therefore all the entries in rest of m columns are 0.