# Determinant of Linear Operator is Well Defined

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## Theorem

Let $V$ be a nontrivial finite dimensional vector space over a field $K$.

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

Then the determinant $\det A$ of $A$ is well defined, that is, does not depend on the choice of a basis of $V$.

## Proof

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

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

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

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

By Change of Coordinate Vectors Under Linear Mapping and since $A_\mathcal B$ and $A_\mathcal C$ represent the same linear operator with respect to different bases, the following diagram commutes: where $u \in V$, and $\left[{u}\right]_\mathcal B$ indicates the coordinate vector of $u$ with repect to $\mathcal B$, and similarly for $\left[{u}\right]_\mathcal C$.

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

Therefore, because a Change of Basis is Invertible we have:

$A_\mathcal B = P^{-1} A_\mathcal C P$

So:

 $\ds \det \left({A_\mathcal B}\right)$ $=$ $\ds \det \left({P^{-1} A_\mathcal C P}\right)$ $\ds$ $=$ $\ds \det \left({P^{-1} }\right) \det \left({A_\mathcal C}\right) \det \left({P}\right)$ Determinant of Matrix Product $\ds$ $=$ $\ds \det \left({P}\right)^{-1} \det \left({A_\mathcal C}\right) \det \left({P}\right)$ Determinant of Inverse Matrix $\ds$ $=$ $\ds \det \left({A_\mathcal C}\right)$

Hence the result.

$\blacksquare$