Determinant of Linear Operator is Well Defined

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 $A_\BB$ and $A_\CC$ be the matrices of $A$ relative to $\BB$ and $\CC$ respectively.

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

We are required to show that $\det A_\BB = \det A_\CC$.

Let $P$ be the change of basis matrix from $\BB$ to $\CC$.

By Change of Coordinate Vectors Under Linear Mapping and since $A_\BB$ and $A_\CC$ represent the same linear operator with respect to different bases, the following diagram commutes:

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where $u \in V$, and $\sqbrk u_\BB$ indicates the coordinate vector of $u$ with respect to $\BB$, and similarly for $\sqbrk u_\CC$.

That is, $P A_\BB = A_\CC P$.

From Change of Basis Matrix is Nonsingular:

$A_\BB = P^{-1} A_\CC P$

So:

\(\ds \map \det {A_\BB}\)

\(=\)

\(\ds \map \det {P^{-1} A_\CC P}\)

\(\ds \)

\(=\)

\(\ds \map \det {P^{-1} } \map \det {A_\CC} \map \det P\)

Determinant of Matrix Product

\(\ds \)

\(=\)

\(\ds \map \det P^{-1} \map \det {A_\CC} \map \det P\)

Determinant of Inverse Matrix

\(\ds \)

\(=\)

\(\ds \map \det {A_\CC}\)

Hence the result.

$\blacksquare$