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How do I see that the representing matrix of the dual map $f^*$ between finite-dimensional dual spaces is given by the transpose of the representing matrix of $f$? Here I want to assume that the matrix $f^*$ is represented with respect to the dual basis.

Apparently this result is very well-known, but I would like to see proof of this.

PinkyWay
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1 Answers1

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Let $f: V \to \bar V$ be a linear transformation represented by the matrix $A = [a^i_j]$ relative to the basis $e_1,\dots,e_n$ for $V$ and $\bar e_1, \dots, \bar e_m$ for $\bar V$. Let $\alpha^1, \dots, \alpha^n$ and $\bar \alpha^1, \dots, \bar \alpha^m$ denote the respective corresponding dual bases.

Suppose that $f^*:\bar V^* \to V^*$ satisfies $f^*(\bar \alpha^i) = \sum_{j} b^i_j \alpha^j$. That is, suppose that $[b^i_j]^T$ is the matrix of $f^*$ relative to the dual bases. It follows that for each $i,j$, we have \begin{align*} b^i_j & = \sum_{k} b^i_k \delta^k_j = \sum_{k} b^i_k \alpha^k e_j = \left(\sum_{k} b^i_k \alpha^k\right) e_j = f^*(\bar \alpha^i) e_j = (f^* \circ \bar \alpha^i) e_j \\ & = (\bar \alpha^i \circ f)(e_j) = \bar \alpha^i(f(e_j)) = \bar \alpha^i\left(\sum_{k}a^k_j \bar e_k\right) = \sum_{k}a^k_j \bar \alpha^i \bar e_k = \sum_{k}a^k_j \delta^i_k = a^i_j \end{align*} as desired. That is, the matrix of $f^*$ with respect to the dual bases is $[b^i_j]^T = [a^i_j]^T = A^T$.

Antonio AN
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Ben Grossmann
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  • Are there any adjustments to be made when $V$ is a complex vector space? – Roland Sep 10 '17 at 19:45
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    @Roland well for one, the bars become a bit confusing. More seriously, this proof only works if you consider the application $\langle \alpha, e \rangle$ to be bilinear. It is more common to consider this application to be sesquilinear, which is to say antilinear with respect to its argument from the dual space. – Ben Grossmann Sep 10 '17 at 19:52
  • You took the matrix of dual space as bij transpose. Suppose we wont take the transpose, then what complications will come? – Avinash Bhawnani Aug 28 '18 at 19:44
  • @AvinashBhawnani The only difference is that instead of ending by stating that $B = A$, I would end by stating that $B^T = A$. – Ben Grossmann Jul 10 '20 at 07:03
  • To anyone who ends up here, I present an equivalent computation (more cleanly, in my opinion) over here. – Ben Grossmann Nov 11 '22 at 20:06