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Let $\lambda$ be an eigenvalue of an n x n matrix Aand let x be an eigenvector belonging to $\lambda$. Show that e$^\lambda$ is an eigenvalue of e$^A$ and x is an eigenvector of e$^A$ belonging to e$^\lambda$.

A study buddy and I were discussing this and we came up with two different approaches, though we're unsure if our work is correct.

My thoughts:

A x = $\lambda$ x

*e*$^A$ = Xe$^D$X$^{-1}$

A = XDX$^{-1}$

Where D is a diagonalizable n x n matrix and X is an invertible n x n matrix.

My attempt was to do this:

*e*$^A$ x = e$^\lambda$ x

*e*$^A$ x - *e*$^\lambda$ x = 0

(*e*$^A$ - *e*$^\lambda$ I) x = 0

(e$^A$ - e$^\lambda$ I) = 0

det(*e*$^A$ - *e*$^\lambda$ I) = 0

det(*e*$^A$) - det(e$^\lambda$ I) = 0

det(e$^A$) = e$^\lambda$

However, this is probably the wrong approach to this problem. Can anyone please give me a tip / help me out here? *e*$^D$ will have diagonals that correspond to A's eigenvalues, yes? Should I start from there instead?

NotAStudentForReal
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  • (1) You don't know that $A$ is diagonalisable (although this is a good way to start looking at the problem) and (2) you can't split $\det$ up like you did above. – copper.hat Nov 19 '15 at 06:50

1 Answers1

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Suppose $Av=\lambda v$. Then $A^k v = \lambda^k v$. Then $e^{A}v = \sum_k {1 \over k!} A^k v = \sum_k {1 \over k!} \lambda^k v = e^\lambda v$.

copper.hat
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  • That was the method my buddy used, I believe. Thank you, by the way. We were scratching our heads for a short while. – NotAStudentForReal Nov 19 '15 at 06:50
  • Oh, just for clarification: that summation is the definition of an exponential matrix, right? We can use it because we're only given one eigenvalue? – NotAStudentForReal Nov 19 '15 at 06:52
  • Yes to the first, $e^A = \sum_k {1 \over k!} A^k$, and I don't understand what you are asking for the second. – copper.hat Nov 19 '15 at 06:55
  • Never mind, I was misreading what my book was saying. Thanks again. – NotAStudentForReal Nov 19 '15 at 07:00
  • Is the converse true? – some_math_guy Sep 24 '23 at 21:40
  • @some_math_guy Not sure what you mean by converse. If $A$ has eigenvalues $0,2 \pi i$ then both will map to $1$, so you cannot take an inverse, if that is what you were asking. – copper.hat Sep 24 '23 at 21:52
  • I meant, if $\exp (A)$ has eigenvalue $\lambda$, then $A$ has eigenvalue $\log(\lambda)$? – some_math_guy Sep 24 '23 at 22:22
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    @some_math_guy No, because $\exp$ is not injective. Both $2 \pi i, 4 \pi i$ map to $1$, but $\log 1 = 0$. – copper.hat Sep 24 '23 at 22:28
  • I was asking because they seem to use that in this question https://math.stackexchange.com/questions/348699/showing-that-the-exponential-map-mathrmexp-mathfraksl2-mathbbr-to-ma, see my comment in the first answer, they say that exp(Y) =(-1,1 ; 0,-1) has a double eigenvalue, but I think that comes from (-1,1 ; 0,-1) having a double eigenvalue, though that implication shouldn't be correct – some_math_guy Sep 26 '23 at 10:45
  • @some_math_guy Concerning the question you referenced, the only solutions to $e^\lambda = -1$ are $(2n+1) \pi i$. If $X$ is real then the eigenvalues must be imaginary (since the trace is zero) and conjugates. If they are non zero, then $X$ must be diagonalisable, which is a contradiction. Hence they are, in this case, both zero. – copper.hat Sep 26 '23 at 21:30