Throughout is a field and all vector spaces are over Recall that the dual space of a vector space is the space of all linear maps It is well-known that if is finite dimensional, then

**Definition**. The **dual** of a linear transformation is the linear transformation defined by for all

**Problem 1**. Let be a linear transformation, where and are finite dimensional. Prove that is an isomorphism if and only if is an isomorphism.

**Solution**. Suppose first that is an isomorphism. In order to prove that is an isomorphism, we only need to show that is onto because, since is an isomorphism, we have So suppose, to the contrary, that is not onto. Then there exists a subspace of such that Choose any and define by where Note that and Clearly for all i.e. which is not possible because and is an isomorphism. Conversely, suppose that is an isomorphism. Then and thus we only need to show that is one-to-one: if then for all and hence, since is onto, we get for all i.e.

**Problem 2**. Let be a linear transformation, where Prove that the characteristic polynomials of and are equal.

**Solution**. Choose a basis for and let be the standard basis for with respect to i.e. for all Here is the Kronecker delta. Suppose that is the matrix of in and is the matrix of in Then

for all and hence

On the other hand, since we have

Now (1) and (2) give us i.e. is the transpose of and we know that the characteristic polynomial of any (square) matrix is the same as the characteristic polynomial of its transpose.