-2
$\begingroup$

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.

Question:

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).

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$
  • 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)

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.