$X$ is a matrix. Let $v$ be an eigenvector of $X$ with corresponding eigenvalue $a$. Show that $v$ is also an eigenvector of $e^{X}$ with eigenvalue $e^{a}$

If $X$ is diagonalizable, then we can start writing out terms using Taylor expansion of $e^{X}$ but I can't seem to get anywhere.

Thanks for the help

Edit: Corrected question to read 'Let $v$ be an eigenvector of $X$' instead of 'Let $v$ be an eigenvector of $e^X$'.

  • $\begingroup$ Did you mean to write, “Let $v$ be an eigenvector of $X$... ?” $\endgroup$
    – amd
    Feb 1, 2016 at 7:18

3 Answers 3


You don't need to assume that $X$ is diagonalisable to use the "Taylor expansion". By definition, $$ \exp(X) = \sum_{k=0}^\infty \frac{1}{k!} X^k $$ Also, if $Xv = av$, then $X^2v = X(Xv) = aXv = a^2v$, etc. By induction, $X^n v = a^n v$.

Hence $$ \exp(X)v = \left(\sum_{k=0}^\infty \frac{1}{k!} X^k \right)v = \sum_{k=0}^\infty \frac{1}{k!} X^k v = \sum_{k=0}^\infty \frac{1}{k!} a^k v = \left( \sum_{k=0}^\infty \frac{1}{k!} a^k \right) v = e^a v. $$

  • $\begingroup$ I was looking at the OP and the previous answer wondering why they were talking about diagonalizability/upper triangular (e.g. Jordan?) form. I thought maybe I missed something. $+1$ for putting my mind at ease! $\endgroup$ Jan 31, 2016 at 22:53
  • 3
    $\begingroup$ Maybe it's just me, but the second equality in the last line seems "non-trivial" to me - specifically, it is implicit from the equality that matrix multiplication commutes with limits involving matrices. Though it is obvious (in that it is easy to prove) that scalars commute with limits, the above fact may not be obvious (especially if the OP is not too familiar with operator norms etc. which would be used to prove such types of statements). In light of this, perhaps you could just add a comment alluding to this fact/theorem that it being used. $\endgroup$
    – John Don
    Mar 31, 2017 at 14:49

If we let $ \Phi(t) = e^{t X}$, we see that $\Phi$ satisfies the (matrix) equation $\dot{Y} = Y X$ subject to the initial condition $Y(0) = I$.

Let $\xi(t) = \Phi(t) v$, where $Xv = a v$, then we see that $\dot{\xi}(t) = \Phi(t) X v = a \Phi(t)v= a \xi(t)$, and so $\xi(t) = e^{a t} v$. Taking $t=1$ we get $e^X v = e^a v$.


Without restricting the generality, we can suppose that the ground field is algebraically closed. There exists $P$ such that $B=PXP^{-1}$ is a sup triangular matrix, and $P(v)$ is an eigenvector of $B$ associated to $a$, $exp(B)=Pexp(A)P^{-1}$, $P(v)$ is an eigenvector of $exp(B)$ associtated to $e^a$. This implies that $v$ is an eigenvector of $exp(X)$ associated to $e^a$.


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