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.

Here is the step-by-step mathematical derivation and logical reasoning:

1. Definition of Matrix Exponential:
For any square matrix A, the matrix exponential $e^A$ is defined by the infinite power series:

$$e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+...=\sum _{n=0}^{\infty }\frac{A^n}{n!}$$

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

2. Eigenvalue and Eigenvector Relation:
We are given that $\lambda$ is an eigenvalue and $v$ is the corresponding eigenvector of matrix $A$. By definition:

$$Av=\lambda v$$

By induction, applying matrix $A$ repeatedly $n$ times to the eigenvector $v$ gives:

$$A^{n}v={\lambda }^{n}v$$

for any non-negative integer $n$.

3. Applying the Matrix Exponential to the Eigenvector:
Now, we multiply the matrix exponential $e^A$ by the vector $v$:

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

Using the linearity of matrix multiplication, we distribute the vector $v$ inside the summation:

$$e^{A}v=\sum _{n=0}^{\infty }\frac{A^{n}v}{n!}$$

Substitute $A^{n}v={\lambda }^{n}v$ into the series:

$$e^{A}v=\sum _{n=0}^{\infty }\frac{{\lambda }^{n}v}{n!}$$

Since $v$ is a constant vector, we can factor it out of the summation:

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

The scalar series inside the parenthesis is the standard Taylor series expansion for the exponential function of a scalar $\lambda$:

$$\sum _{n=0}^{\infty }\frac{{\lambda }^n}{n!}=e^{\lambda }$$

Therefore, we obtain:

$$e^{A}v=e^{\lambda }v$$

Conclusion:
From this final equation, we can see that:
• $e^{\lambda }$ acts as the eigenvalue of the matrix $e^A$ (proving Statement 1 is correct).
• The corresponding eigenvector is $v$ (proving Statement 2 is correct).