Let $A \in Mat_{n,n}(\mathbb C)$.

Let $g_A(t)$ denote $\exp(tA)$ and $\exp(A)$ denote the matrix $g_A(1)$.

Let $\alpha \in \mathbb R$ and $B = \alpha A$.

I've shown $\exp(B) = g_A(\alpha)$ by just substituting $B$ with $\alpha A$ (is this correct?).

I've shown that $\exp(A)$ is invertibel by considering $\exp(-tA)$ and considering the derivative of product $\exp(-tA) \exp(tA)$ which is constant equal to $I_n$ (an easier way to do this?).

I've shown that $\exp(0) = I_n$ by using $A = 0$ and $\alpha = 0$ and we know that $\exp(0A) = I_n$ by definition.


However I cannot show that $\exp(A)$ is a real matrix if $A$ is a real matrix ? (I should relate the question to solutions of linear differential equations).

Now don't assume $A$ is real.

Also if $C$ is a matrix that commutes with $A$, that is $AC = CA$, I can show that $C$ commutes with $\exp(A)$ by using Putzer's algorithm.

But how can I show $\exp(A) \exp(B) = \exp(A+B)$ ? (Again I should relate the question to solutions of linear differential equations).

  • $\begingroup$ Since you asked why $\exp A$ is invertible , notice that you can prove that $\det \exp A= \exp (tr(A))$ $\endgroup$ – Gabriel Romon May 16 '14 at 17:39
  • Let $X$ be an $n×n$ real or complex matrix. The exponential of $X$, denoted by $e^X$ or $\exp(X)$, is the $n×n$ matrix given by the power series: $$e^X = \sum_{k=0}^\infty{1 \over k!}X^k.$$ From this definition, you can see that the exponential of a real matrix is a real matrix since $X^k$ is a real matrix.

  • For $\exp(A)\exp(B)=\exp(A+B)$ see this (Theorem 5, p. 4)

  • $\begingroup$ Do you know Putzer's algorithm for constructing $e^X$ ? In this algorithm we use the eigenvalues of $X$. How do I know the matrix function constructed is real from this ? $\endgroup$ – Shuzheng May 16 '14 at 16:16
  • $\begingroup$ If a matrix $A$ is a real matrix it can have complex eigenvalues ? We assume $A$ is complex matrix with real entries and must prove $e^A$ is real ? $\endgroup$ – Shuzheng May 16 '14 at 16:38
  • $\begingroup$ No. No. If $A$ is a real matrix then $A$ can have complex eigenvalues. See this. Sorry I do not know how to prove that $e^A$ is real using Putzer's algorithm. $\endgroup$ – Jika May 16 '14 at 16:50

$\exp(A)$ is real when $A$ is real through the power series expansion.

  • $\begingroup$ How about Putzers algorithmn ? $\endgroup$ – Shuzheng May 22 '14 at 6:06

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