---
tag: 0ZPB3180AY
kind: statement
role: proposition
name: Matrix of the dual map is the transpose
msc: [15A04]
uses: [0C2TQMW5A1, 0W6M3PSD08, 0YYMY95FWS, 0W809K6AW0]
sources:
  - url: https://en.wikipedia.org/wiki/Transpose_of_a_linear_map
    locator: Representation as a matrix
---

## Statement

Let $F$ be a field, and let $V$ and $W$ be finite-dimensional vector spaces over $F$.
Let $B$ be a basis of $V$ and $C$ a basis of $W$, with [[dual bases|0ZD2CF6ZS0]] $B^*$ and $C^*$.
Let $T \colon V \to W$ be a linear map with [[dual map|0AZ5ARP287]] $T^*$.
Then
$$[T^*]_{B^*, C^*} = \left([T]_{C, B}\right)^T,$$
where $[\,\cdot\,]$ is the [[matrix of a linear map|0YQSGNYE3K]] and the exponent $T$ is the [[transpose|09DXWYRD43]].

## Remarks

The proof computes entry $(j, i)$ of $[T^*]$ as $T^*(c_i^*)(b_j) = c_i^*(T(b_j))$, which is entry $(i, j)$ of $[T]$.
With the [[annihilator|0ASHNYF7AG]], this gives a proof that [[row rank equals column rank|0TZAG60ZYT]].

## Explanations

### L1 human

The dual map is the linear algebra form of the transpose of a matrix.
In dual bases, its matrix is the transpose of the matrix of the original map.
This explains why rows and columns play symmetric roles.