Exactly what it says on the title. How do I show that e $^A$ is nonsingular for any diagonalizable matrix A?
I know that e $^A$ = X e $^D$X$^{-1}$
and
A = XDX$^{-1}$
where A, X, and D are all n x n matrices. Since A is diagonalizable, I don't need to use the definition of an exponential matrix. Where should I begin with proving that e $^A$ is nonsingular?
On
The simplest way to do this using your approach is to just see that if $$A=XDX^{-1}$$ then letting $$D=\begin{pmatrix}\alpha_1 & 0 & 0 & \ldots \\ 0 & \alpha_2 & 0 & \ldots \\ 0 & 0 & \alpha_3 & \ldots \\ \vdots & \vdots & \vdots & \ddots\end{pmatrix}$$ we have $$e^D=\begin{pmatrix}e^{\alpha_1} & 0 & 0 & \ldots \\ 0 & e^{\alpha_2} & 0 & \ldots \\ 0 & 0 & e^{\alpha_3} & \ldots \\ \vdots & \vdots & \vdots & \ddots\end{pmatrix}$$ which isn't singular, hence neither is $$e^A=Xe^DX^{-1}$$
Somewhat more directly would be to note that if $Av=\lambda v$ then $e^Av=e^{\lambda}v$ and, since the eigenvectors of a diagonalizable matrix must span the space and are also eigenvectors of $e^A$ with non-zero eigenvalues and span the space. Thus, there can be no eigenvector with eigenvalue $0$ as it would have to be linearly independent of the eigenvalues of $A$. The intuition here is that matrix exponentiation can basically just be applied to the eigenvalues.
By the fact that $$\exp(A) = X \exp(D) X^{-1},$$ we have \begin{equation} \begin{split} \det(\exp(A)) & = \det(X \exp(D) X^{-1}) \\ & = \det(X) \det(\exp(D)) \det(X^{-1}) \\ & = \det(X) \prod_{i=1}^n e^{\lambda_i} \det(X)^{-1} \\ & = \prod_{i=1}^n e^{\lambda_i} \\ & = e^{\sum_{i=1}^n \lambda_i} \\ & = e^{\text{tr}(D)} \\ & \neq 0. \end{split} \end{equation}
Since $\exp(A)$ has a nonzero determinant, it is nonsingular.
Thanks to Thomas' reminder, we remark that $\text{tr}(D)$ is a complex-valued and $\det(\exp(A))$ is a nonzero complex number. In particular, if $\text{tr}(D)$ is real-valued, then we also see that $\det(\exp(A))$ is a positive real number.