Assume {v1, . . . , vn} is a basis of a vector space V , and T : V > W is an isomorphism where W is another vector space. Show that the vectors w1 = T(v1), . . . ,wn = T(vn) are a basis of W.
This statement becomes false if we drop the assumption that T is invertible. Demonstrate this by finding a counterexample.

Step-by-step explanation:

To prove that form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set is linearly independent if and only if  the equation

implies that

.

Recall that for i=1,...,n. Consider to be the inverse transformation of T. Consider the equation

If we apply to this equation, then, we get

Since T is linear, its inverse is also linear, hence

which is equivalent to the equation

Since are linearly independt, this implies that , so the set is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist such that

Since T is surjective, there exists a vector v in V such that T(v) = w. Since is a basis of v, there exist , such that

Then, applying T on both sides, we have that

which proves that generate the whole space W. Hence, the set is a basis of W.

Consider the linear transformation , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of

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