e^A denotes the exponential of a square matrix A. Suppose λ is an eigenvalue and v is the corresponding eigen-vector of matrix A.

Consider the following two statements :

Statement 1 : eλ is an eigenvalue of e^A .

Statement 2 : v is an eigen-vector of e^A .

Which one of the following options is correct?

The correct option is: Both the statements are correct.

Let us analyze the definition of the matrix exponential and the properties of eigenvalues and eigenvectors to verify both statements.

For a square matrix $A$, the matrix exponential $e^A$ is defined by the infinite Taylor series expansion:

$e^A=I+A+\frac{1}{2!}{A}^2+\frac{1}{3!}{A}^3+\cdots =\sum _{k=0}^{\infty }\frac{1}{k!}{A}^k$

where $I$ is the identity matrix of the same size as $A$, and $A^0=I$.

We are given that $\lambda$ is an eigenvalue of $A$ corresponding to the eigenvector $v$. By definition, this means:
$Av=\lambda v$ (with $v\ne 0$)

Let us determine the effect of higher powers of $A$ on the eigenvector $v$.
For $k=2$:
$A^{2}v=A\left(Av\right)=A\left(\lambda v\right)=\lambda \left(Av\right)={\lambda }^{2}v$
By mathematical induction, for any non-negative integer $k$, we have:
$A^{k}v={\lambda }^{k}v$

Now, we apply the matrix exponential $e^A$ to the eigenvector $v$:

$e^{A}v=\left(\sum _{k=0}^{\infty }\frac{1}{k!}{A}^k\right)v$

Using the linearity of matrix multiplication: