Suppose B is diagonalizable, then $e^{B}$ is diagonalizable and $e^{\lambda}$ is an eigenvalue of $e^{B}$ if and only if $\lambda$ is an eigenval of B

Question

As stated above, Suppose B is diagonalizable, then $e^{B}$ is diagonalizable and $e^{\lambda}$ is an eigenvalue of $e^{B}$ if and only if $\lambda$ is an eigenvalue of B. I'm struggling to determine if this is true or not, is there proof or theorem to justify this one way or the other?

Answer 1

I'll assume $B$ is a $n\times n$ matrix with complex entries. If $B$ is diagonalizable, then we might as well diagonalize it, so we will work in a basis where $B$ is a diagonal matrix. As you probably know, the exponential of $B$ may be defined by the power series $$e^B=\sum_{k=0}^{\infty}\frac{1}{k!}B^k=I + B + \frac{B^2}{2!}+(...) ,$$ where $I$ is appropriate Identity matrix. The identity is of course diagonal and, since the product of diagonal matrices is also diagonal, so is every other term in the series ($B^2$ is diagonal. If you assume $B^k$ is diagonal, then $B^{k+1}$ will be as well). Thus the exponential matrix is a sum of diagonal matrices, and hence diagonal in the same basis. We can say a bit more. There's a theorem in linear algebra which states that an operator is diagonalizable if and only if it has $n$ linearly independent eigenvectors. So we may write $B=\rm{diag}(\lambda_1,...\lambda_n)$, where the diagonal entries are eigenvalues of $B$ (not necessarily distinct or real). Let $v$ be an eigenvector of $B$. We have $Bv=\lambda v$ for some $\lambda$. By the definition presented above, we have $$e^{B}v=\sum_{k=0}^{\infty}\frac{1}{k!}B^k v= \left(1 + \lambda+ \frac{1}{2}\lambda^2+... \right)v=e^{\lambda}v,$$ where I used the fact that $B^k v=\lambda B^{k-1}v=...=\lambda^k v$. So $e^{\lambda}$ in indeed an eigenvalue of $e^B$. The converse, however, is not true. In fact, the exponential is periodic, so $e^{\lambda}=e^{(\lambda + 2\pi iN)}$, for any integer $N$. Thus $e^{(\lambda + 2\pi iN)}$ is an eigenvalue of $e^B$ , but of course that doesn't imply $\lambda + 2\pi iN$ is an eigenvalue of $B$.

Answer 2

Let $B$ be diagonalised as $B=UDU^*$, $U$ is a unitary matrix and $D=diag(\lambda_1, \lambda_2, \cdots, \lambda_n)$. Then Note that $e^B=e^{UDU^*}=Ue^DU^*$ and $e^D=diag(e^{\lambda_1}, e^{\lambda_2}, \cdots, e^{\lambda_n})$ via the expansion $e^A=\sum_{i=0}^\infty \frac{A^i}{i!}$. Now since $e^B=U \begin{pmatrix} e^{\lambda_1} & 0 & \cdots & 0\\ 0 & e^{\lambda_2} & \cdots & 0 \\ & \cdots & \\ 0 & 0 & \cdots & e^{\lambda_n} \end{pmatrix} U^*=U \begin{pmatrix} e^{\lambda_1+2m_1 \pi i} & 0 & \cdots & 0\\ 0 & e^{\lambda_2+2m_2 \pi i} & \cdots & 0 \\ & \cdots & \\ 0 & 0 & \cdots & e^{\lambda_n+2m_n\pi i} \end{pmatrix} U^*$ for positive integers $m_1, m_2, \cdots, m_n$, you can see that only "$\Rightarrow$" is true.