Alternative definition of matrix exponential

Question

Suppose we have a matrix $X$. In some research papers matrix exponential is calculated as $V \times diag(exp(diag(D)))/V$ where $V = $ Eigenvectors of $X$ and $D$ = Eigenvalues of $X$. $diag$ means diagonalizing the matrix. $exp =$ element by element exponential. How is this definition of matrix exponential useful rather than directly exponentiating every element of matrix $X$. Matlab also uses this definition of matrix exponential by its function $expm$.

Answer 1

For a matrix $A\in\mathbb{C}^{n\times n}$ the matrix exponential is defined as $$\tag{1}\label{expA} \exp(A) = \sum_{k=0}^\infty \frac{A^k}{k!} = I + A + A^2/2 + \ldots. $$ The matrix exponential plays a prominent role for systems of ODEs: For an autonomous linear ODE system $$\tag{2}\label{ODEA} y'=A y~~~\text{with}~~~y=y(t)\in\mathbb{C}^n,~y(0)=y_0 $$ the solution at a time $t>0$ is given by the exponential of $tA$ applied to the initial vector $$ y(t)=\exp(tA) y_0. $$

( Applying the exponential on each matrix entry does not solve an ODE system in general.)

Remark: When $A$ is a diagonalizable matrix with eigenvalues $\lambda_1,\ldots,\lambda_n\in\mathbb{C}$ and eigenbasis $V\in\mathbb{C}^{n\times n}$, i.e., $A=V \text{diag}(\lambda_1,\ldots,\lambda_n) V^{-1}$, the matrix exponential $\exp(A)$ as defined in \eqref{expA} can be written as $$ exp(A) = V \text{diag}(\exp(\lambda_1),\ldots,\exp(\lambda_n)) V^{-1}. $$ Make use of the eigendecomposition of $A$ in \eqref{ODEA} to show that $\exp(tA)$ yields a solution of the ODE system.