Proof that the determinant of a matrix exponential equals the exponential of trace of the matrix

Question

The question arises from this tweet:

Answer 1

It was supposed to be a question, but I kept going...

I believe that the idea would be to use the diagonalization of the square matrix $M,$ such as

$$M= P\Lambda P^{-1}$$

where $$\Lambda =\begin{bmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\cdots&\ddots&\vdots\\0&\cdots&\cdots&\lambda_n\end{bmatrix}$$

is the matrix of eigenvalues.

Now since

$$\begin{align} e^M &= \sum_{n=0}^\infty \frac{M^n}{n!}\\ &= \sum_{n=0}^\infty \frac{P\Lambda^n P^{-1}}{n!}\\ &= P\left(\sum_{n=0}^\infty \frac{\Lambda^n }{n!}\right)P^{-1}\\ &=Pe^\Lambda P^{-1} \end{align}$$

The determinant will be

$$\begin{align} \det(e^M) & =\det( P) \det \left( e^\Lambda \right )\det \left( P^{-1}\right)\\ &= \det \left( e^\Lambda \right )\\ &= e^{\lambda_1} e^{\lambda_2}\cdots e^{\lambda_n}\\ &=e^{\lambda_1 + \lambda_2+\cdots +\lambda_n} \end{align}$$

The right side of the equation would be

$$\begin{align} e^{\text{Tr} (M)} & = e^{\text{Tr} \left({P\Lambda P^{-1}}\right)}\\ &=e^{\text{Tr} {(\Lambda)} }\\ &=e^{\lambda_1 + \lambda_2+\cdots +\lambda_n} \end{align} $$