10  basic properties

10.1 \det(\exp(A))=\exp(\text{tr}(A))

We wish to prove that for a diagonalizable matrix A, the determinant of its exponential equals the exponential of its trace:

\det(\exp(A)) = \exp(\text{tr}(A))

We start with the definition of the exponential function for a scalar x:

\exp(x) = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots

We will extend this definition to include the exponential of a matrix A:

\exp(A) = \sum_{n=0}^{\infty}\frac{A^n}{n!} = 1 + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots

If A is diagonalizable:

A = P D P^{-1},

where P is a matrix whose column vectors are the eigenvectors of A, and D is a diagonal matrix whose diagonal elements are the corresponding eigenvalues.

Let’s use the equation above in the definition of the exponential:

\begin{align*} \exp(A) &= \exp\left(P D P^{-1}\right) \\ &= 1 + P D P^{-1} + \frac{1}{2!}\left(P D \cancel{P^{-1}}\right)\left(\cancel{P} D P^{-1}\right) + \ldots \\ &= P D^2 P^{-1} + P D P^{-1} + \frac{1}{2!}P D^2 P^{-1} + \frac{1}{3!}P D^3 P^{-1} + \ldots \end{align*}

If we left-multiply the last expression by P^{-1} and then right-multiply it by P, we get

P^{-1}\exp(A)P = 1 + D + \frac{D^2}{2!} + \frac{D^3}{3!} = \exp(D)

and then reversing the left- and right-multiplications we finally have:

\exp(A) = P \exp(D) P^{-1}.

We are almost there! Let’s take the determinant of both sides. In the step below we will use the following property of determinants: \det(XY)=\det(X)\det(Y). This can be untuitively justified by interpreting the determinant as the rescaling of a parallelogram defined by the linear transformation of a matrix. The rescaled parallelogram after the operation XY is scaled first by Y and then by X, giving the expression we need. Continuing:

\begin{align*} \det(\exp(A)) &= \det(P \exp(D) P^{-1}) \\ &= \det(P)\det(\exp(D))\det(P^{-1}) \\ &= \cancel{\det(P)}\det(\exp(D))\cancel{\det(P^{-1})} \\ &= e^{d_1}e^{d_2}e^{d_3}\ldots \\ &= \exp(d_1+d_2+d_3+\ldots) \\ &= \exp(\text{tr}(D)) \end{align*}

Now, all we have to do is to replace the trace of D by the trace of A. Using the property of the trace of a product: \text{tr}(XY)=\text{tr}(YX):

\text{tr}(A) = \text{tr}(P D P^{-1}) = \text{tr}(\cancel{P^{-1}P} D) = \text{tr}(D).

Finally:

\det(\exp(A)) = \exp(\text{tr}(A))

and that concludes the proof. \blacksquare