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.

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 where T is the corresponding lineat transformation from V to W.

To prove the requirment let be a basis of dim Ker(T).

Now extend this basis of V to n such that, new basis is of the form . Then basis of range T is of the form, which are linearly independent. Therefore according to rank-nullity theorem dim range T=n.

Now extending basis of range space into :

with respect to the basis (note the reverse order of vectors) of V, the matrix of T has the desired form.

Since for any we have all the entries in the first n-column 0, except the entries in row i, column i, equal to 1 for =dim range (T).

for any

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

answer is more than one

step-by-step explanation: