0

If $u$ is an eigen vector of $\exp(A)$ then $u$ is an eigen vector (e.vector) of $A$.

Is it true for a $2 \times 2$ real-valued matrix $A$.

My try: $\exp(A)$ is invertible so all e.values (eigenvalues) are non zero so we can write $$\exp(A)=U(e^aI+N)U^{-1}$$ where $e^a$ is an e.value corresponding to the e.vector $u$. Now I can write $$\exp(A)=U(e^aI+N)U^{-1}=U(e^aI(I+e^{-a}N)U^{-1}$$

So, even if I write $$\log(e^aI)+\log(I+e^{-a}N)=\log(e^aI)-e^{-a}N$$

How will I manage $U$ and $U^{-1}$. Help me here.

Btw, for complex case the statement is not correct. We can get a contradiction here

Let $$A = \begin{bmatrix} 4\pi i & 0\\ 0 & 2\pi i\\ \end{bmatrix}.$$

Then $\exp(A) = Id$ and $v = \begin{bmatrix} 1\\ 1 \end{bmatrix}$ is an eigenvector of $\exp(A)$ but not of $A$.

Look here.

Gary
  • 31,845
Ri-Li
  • 9,038

1 Answers1

4

This is not true: If $A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ then $A^{2n} = (-1)^nI_2$ and $A^{2n+1} = (-1)^n A$ so that $$ \exp(\pi A) = I_2 \cos(\pi) + A \sin(\pi) = - I_2. $$ Any nonzero vector is eigenvector of $\exp(\pi A)$, but not of $A$.

jlewk
  • 1,867
  • Indeed. Thanks. I fixed it. – jlewk Oct 24 '22 at 02:21
  • 1
    I meant to identify the complex 1 with $I_2$ and the complex $i=\sqrt{-1}$ with $A$, so that $A^2=-I_2$ and the calculation $\exp(\pi A)=-I_2$ is equivalent to $e^{i\pi}=-1$. I believe the current choice works out. Thanks for spotting the initial typo. – jlewk Oct 24 '22 at 03:23
  • Oh yes, I misread the exponents. – lonza leggiera Oct 24 '22 at 06:11