---
tag: YTKK8764G7
kind: proof
name: Proof that the determinant of a linear map does not depend on the basis
msc: [15A04, 15A15]
proves: 0KMN32XGPF
uses: [0C3CFHDN6R, 03F5KP94K0, 0YYMY95FWS, 0MQXVZEBMT, 0KWXYZPJKN, 04CPJ6AF00, 0TWTS88TWN, 0Z4BWARC38]
sources:
  - url: https://en.wikipedia.org/wiki/Determinant
    locator: Determinant of an endomorphism
---

## Proof

We show that every basis gives the same value of $\det [T]_B$. Then the definition of $\det T$ gives this value for each basis.

1. Let $n = \dim V$. The [[determinant|0C3CFHDN6R]] is defined for $n \times n$ matrices with $n \ge 1$, so $n \ge 1$ here.
2. Let $B$ and $B'$ be two bases of $V$. Let $[T]_B$ and $[T]_{B'}$ be the [[matrices of $T$|0YYMY95FWS]] in these bases.
3. Let $P$ be the [[change of basis matrix|0KWXYZPJKN]] from $B$ to $B'$. By the [[change of basis formula|0MQXVZEBMT]], $P$ is invertible and $[T]_{B'} = P^{-1} [T]_B P$.
4. By the [[product rule|04CPJ6AF00]], $\det [T]_{B'} = \det(P^{-1}) \det [T]_B \det P$.
5. The ring $F$ is commutative, so the right side equals $\det(P^{-1}) \det P \cdot \det [T]_B$.
6. By the product rule, the definition of the [[inverse|0TWTS88TWN]] and the [[determinant of the identity|0Z4BWARC38]], $\det(P^{-1}) \det P = \det(P^{-1} P) = \det I_n = 1$.
7. By steps 4 to 6, $\det [T]_{B'} = \det [T]_B$ for any two bases $B$ and $B'$.
8. The [[determinant of $T$|03F5KP94K0]] is defined as $\det [T]_{B'}$ for a chosen basis $B'$. By step 7, $\det T = \det [T]_B$ for the given basis $B$.

## Idea

Changing the basis replaces the matrix $[T]_B$ by a similar matrix $P^{-1} [T]_B P$. The determinant is multiplicative, so the factors $\det P^{-1}$ and $\det P$ cancel.