## Dual of a linear transformation

Posted: January 9, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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. $\Box$