Determinant of Linear Transformation 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:

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

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \operatorname{det}(A_{\mathcal B})\)	\(=\)	\(\displaystyle \operatorname{det}(P^{-1}A_{\mathcal C}P)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \operatorname{det}(P^{-1})\operatorname{det}(A_{\mathcal C})\operatorname{det}(P)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Determinant of Matrix Product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \operatorname{det}(P)^{-1}\operatorname{det}(A_{\mathcal C})\operatorname{det}(P)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Determinant of Inverse
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \operatorname{det}(A_{\mathcal C})\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

This concludes the proof.

$\blacksquare$