The Determinant of a Function of Matrices

Functions of matrices are defined by their Taylor series expansions; for example, for some matrix \(\mathsf{M}\) $$ e^\mathsf{M}=\mathsf{I+M}+\frac{1}{2}\mathsf{M}^2+\frac{1}{3!}\mathsf{M}^3+\dots $$

Theorem If \(\mathsf{M}\) is diagonalisable, then $$ \mathrm{det}(e^\mathsf{M})=e^{\mathsf{Tr(M)}} $$

Proof

Let \(\mathbf{v}\) be an eigenvector of \(e^\mathsf{M}\). $$ e^\mathsf{M}\mathbf{v} =(\mathsf{I+M}+\frac{1}{2}\mathsf{M}^2+\frac{1}{3!}\mathsf{M}^3+\dots)\mathbf{v} =\lambda\mathbf{v} $$ where \(\lambda\) is the corresponding eigenvalue.

If \(\mathbf{v}\) is also an eigenvector of \(\mathsf{M}\), then $$ \mathsf{M}\mathbf{v}=\mu\mathbf{v} $$ $$ \begin{align} e^{\mathsf{M}}\mathbf{v} &=(\mathsf{I+M}+\frac{1}{2}\mathsf{M}^2+\frac{1}{3!}\mathsf{M}^3+\dots)\mathbf{v} \\ &=\mathbf{v}+\mathsf{M}\mathbf{v}+\frac{1}{2}\mathsf{M}^2\mathbf{v}+\frac{1}{3!}\mathsf{M}^3\mathbf{v}+\dots \\ &=\mathbf{v}+\mu\mathbf{v}+\frac{1}{2}\mu^2\mathbf{v}+\frac{1}{3!}\mu^3\mathbf{v}+\dots \\ &=\mathbf{v}(1+\mu+\frac{1}{2}\mu^2+\frac{1}{3!}\mu^3+\dots) \\ &=e^{\mu}\mathbf{v} \end{align} $$ Hence, $$ \lambda\mathbf{v}=e^{\mu}\mathbf{v} \Rightarrow \lambda=e^{\mu} $$

As this result applies to any other eigenvector \(\mathbf{v}_n\) with eigenvalue \(\lambda_n\) for \(e^\mathsf{M}\) and eigenvalue \(\mu_n\) for \(\mathsf{M}\), we have $$ \begin{align} \mathrm{det}(e^\mathsf{M})&=\lambda_1\lambda_1\dots\lambda_n \\ &=e^{\mu_1}e^{\mu_2}{\dots}e^{\mu_n} \\ &=e^{\mu_1+\mu_2+\dots+\mu_n} \\ &=e^{\mathsf{Tr(M)}} \end{align} $$

Therefore, $$ \bbox[7pt, border: 1px solid black] { \mathrm{det}(e^\mathsf{M})=e^{\mathsf{Tr(M)}} } $$

Now we are left to prove that any eigenvector \(\mathbf{v}_n\) of \(e^\mathsf{M}\) is also an eigenvector of \(\mathsf{M}\). We will use the following theorem [1]:

Theorem If two diagonalisable matrices commute then they are simultaneously diagonalisable.

If we have some similiarity matrix \(\mathsf{S}\) that diagonalises \(\mathsf{M}\), could it diagonalise \(\mathsf{M}^n\) as well?

i.e. \(\mathsf{SMS}^{-1}=\mathsf{D_1}\) and \(\mathsf{SM}^n\mathsf{S}^{-1}=\mathsf{D_n} \qquad (1)\)

Evaluating the commutator of \(\mathsf{M}\) and \(\mathsf{M}^n\), $$ [\mathsf{M,M}^n]=\mathsf{MM}^n-\mathsf{M}^{n}\mathsf{M}=\mathsf{M}^{n+1}-\mathsf{M}^{n+1}=\mathsf{0} $$ So \((1)\) is satisfied.

Applying \(\mathsf{S}\) to \(e^\mathsf{M}\), $$ \begin{align} \mathsf{S}e^{\mathsf{M}}\mathsf{S}^{-1} &=\mathsf{S}(\mathsf{I+M}+\frac{1}{2}\mathsf{M}^2+\frac{1}{3!}\mathsf{M}^3+\dots)\mathsf{S}^{-1} \\ &=\mathsf{SIS}^{-1}+\mathsf{SMS}^{-1}+\frac{1}{2}\mathsf{SM}^{2}\mathsf{S}^{-1}+\frac{1}{3!}\mathsf{SM}^{3}\mathsf{S}^{-1}+\dots \\ &=\mathsf{I+D_1}+\frac{1}{2}\mathsf{D_2}+\frac{1}{3!}\mathsf{D_3}+\dots \end{align} $$ which is another diagonal matrix. Hence \(\mathsf{S}\) also diagonalises \(e^\mathsf{M}\). Noting that the matrix \(\mathsf{S}^{-1}=(\mathbf{v}_1 \ \mathbf{v}_1 \dots \mathbf{v}_n)\) has columns \(\mathbf{v}_n\) that are the eigenvectors of \(\mathsf{M}\) (in order to diagonalise \(\mathsf{M}\)), we conclude that they are also the eigenvectors of \(e^\mathsf{M}\).

Therefore, \(e^\mathsf{M}\) and \(\mathsf{M}\) have the same eigenvectors.

The proof is complete.