Determinant of Linear Operator is Well Defined

Theorem
Let $V$ be a vector space over a field $K$.

Let $A : V \to V$ be a linear transformation of $V$.

Then the determinant $\operatorname{det}A$ of $A$ is well defined.

Proof
Let $A_{\mathcal B}$ and $A_{\mathcal C}$ be the matrices of $A$ relative to $\mathcal B$ and $\mathcal C$ respectively.

Let $\operatorname{det}$ also denote the determinant of a matrix.

We are required to show that $\operatorname{det}A_{\mathcal B} = \operatorname{det}A_{\mathcal C}$.

Let $P$ be the change of basis matrix from $\mathcal B$ to $\mathcal C$.

Since $A_{\mathcal B}$ and $A_{\mathcal C}$ represent the same linear transformation with respect to different bases, the following diagram commutes:


 * Determinant Independent of Basis.png

Where $u \in V$ is some vector, and $[u]_{\mathcal B}$ indicates the representation of $u$ with repect to $\mathcal B$, and similarly for $[u]_{\mathcal C}$.

That is, $PA_{\mathcal B} = A_{\mathcal C}P$.

Therefore, because a Change of Basis is Invertible we have $A_{\mathcal B} = P^{-1}A_{\mathcal C}P$. So:

This concludes the proof.