Recipe

Find the matrix of a dual map

Use when: you know T's matrix relative to a basis and its dual basis, and need T^{*}'s matrix or its exact effect on one given functional

Goal: the matrix of T^{*} relative to the dual bases, and the explicit coordinates or formula for T^{*} applied to any functional you're given

Taught in The matrix of the dual map is the transpose

1. 1

   State T's matrix A exactly as given, making sure it's relative to the same bases the problem specifies for both V and W.

   Example

   T has matrix A=\begin{pmatrix}1&2\\3&1\end{pmatrix} relative to the standard basis of \mathbb R^2 (and its dual).

   Why: this section's whole claim is about T^{*}'s matrix relative to the DUAL of these exact bases, so the wrong starting matrix produces the wrong transpose.

2. 2

   Flip A across its main diagonal: the entry in row i, column j moves to row j, column i.

   Example

   A^{\mathsf T}=\begin{pmatrix}1&3\\2&1\end{pmatrix}, since the 2 in row 1 moves to row 2 and the 3 in row 2 moves to row 1.

   Why: this section proves T^{*}'s matrix relative to the dual bases is always exactly A^{\mathsf T}, never A itself or its inverse.

3. 3

   Express the functional \psi as coordinates relative to the dual basis of W^{*}, the same way any vector gets coordinates relative to a basis.

   Example

   \psi(x,y)=2x-y has coordinates (2,-1) relative to the dual of the standard basis, since \psi=2\gamma_1-\gamma_2.

   Why: T^{*}'s matrix multiplies coordinate vectors, so \psi needs to be one before the matrix from Step 2 can act on it.

4. 4

   Carry out the ordinary matrix-vector product to get T^{*}(\psi)'s coordinates relative to the dual basis of V^{*}.

   Example

   A^{\mathsf T}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}1(2)+3(-1)\\2(2)+1(-1)\end{pmatrix}=\begin{pmatrix}-1\\3\end{pmatrix}

   Why: this single multiplication is the entire content of "T^{*}'s matrix acting on \psi," once both are written as coordinates.

5. 5

   Read the output coordinates back into an explicit functional, then compare against computing T^{*}(\psi)=\psi\circ T directly from the definition.

   • the two methods agree — the transpose computation is confirmed correct.
   • the two methods disagree — recheck which bases A was built relative to, since a mismatched basis is the most common source of error here.

   Why: T^{*}(\psi)=\psi\circ T can also be computed straight from the definition, so agreement between the two routes is genuine evidence, not the same arithmetic checked twice.