Dual of a linear transformation

Posted: January 9, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , ,

Throughout F is a field and all vector spaces are over F. Recall that the dual space V^* of a vector space V is the space of all linear maps V \longrightarrow F. It is well-known that if V is finite dimensional, then \dim_F V = \dim_F V^*.

Definition. The dual of a linear transformation f: V \longrightarrow W is the linear transformation f^*: W^* \longrightarrow V^* defined by f^*(a)=af, for all a \in W^*.

Problem 1. Let f : V \longrightarrow W be a linear transformation, where  V and W are finite dimensional. Prove that f is an isomorphism if and only if f^* is an isomorphism.

Solution. Suppose first that f^* is an isomorphism. In order to prove that f is an isomorphism, we only need to show that f is onto because, since f^* is an isomorphism, we have \dim V = \dim V^*=\dim W^*=\dim W. So suppose, to the contrary, that f is not onto. Then there exists a subspace (0) \neq W_0 of W such that W=f(V) \oplus W_0. Choose any 0 \neq a_0 \in W_0^* and define a \in W^* by a(w)=a_0(w_0), \ w \in W, where w=f(v)+w_0. Note that a \neq 0 and a \in W^*. Clearly a(f(v))=0 for all v \in V, i.e. f^*(a)=0, which is not possible because a \neq 0 and f^* is an isomorphism. Conversely, suppose that f is an isomorphism. Then \dim V = \dim W and thus we only need to show that f^* is one-to-one: if 0=f^*(a)=af, then a(f(v))=0 for all v \in V and hence, since f is onto, we get a(w)=0 for all w \in W, i.e. a=0. \ \Box

Problem 2. Let f : V \longrightarrow V be a linear transformation, where \dim_F V = n < \infty. Prove that the characteristic polynomials of f and f^* are equal.

Solution. Choose a basis \mathcal{B}=\{v_1, \cdots , v_n \} for V and let \mathcal{B}^*=\{e_1, \cdots , e_n \} be the standard basis for V^* with respect to \mathcal{B}, i.e. e_i(v_j)=\delta_{ij}, for all 1 \leq i,j \leq n. Here \delta_{ij} is the Kronecker delta. Suppose that X =[x_{ij}] is the matrix of f in \mathcal{B} and Y=[y_{ij}] is the matrix of f^* in \mathcal{B}^*. Then

f^*(e_j)=e_jf = \sum_{k=1}^n y_{kj}e_k,

for all j, and hence

f^*(e_j)(v_i)=\sum_{k=1}^n y_{kj}e_k(v_i)=y_{ij}. \ \ \ \ \ \ \ \ \ \ \ \ (1)

On the other hand, since f(v_i) = \sum_{k=1}^n x_{ki}v_k, we have

f^*(e_j)(v_i)=e_jf(v_i)=\sum_{k=1}^n x_{ki}e_j(v_k)=x_{ji}. \ \ \ \ \ \ \ \ \ \ \ \ \ (2).

Now (1) and (2) give us x_{ji}=y_{ij}, i.e. Y is the transpose of X and we know that the characteristic polynomial of any (square) matrix is the same as the characteristic polynomial of its transpose. \Box

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