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$e^X$ for matrix $X$ is defined as an always-converging taylor series (provided that $X$ is a $n \times n $ complex matrix): $$e^X:=\sum_{k=0}^{\infty}\frac{X^k}{k!} $$ A thought occurred to me that we might as well define $\cos(X):=\frac12(e^{iX}+e^{-iX})$ and $\sin(X):=\frac1{2i}(e^{iX}-e^{-iX})$. Now some obvious questions arise:

  • Is there a generalization for $2\pi$, the period of sine and cosine? Perhaps the best way to do so is to generalize the Euler's Identity $e^{2i\pi}=1$; Is there matrix $T$ such that $e^T=1$? This implies that $\cos (X+T)=\cos (X), \sin(X+T)=\sin (X)$.
  • A simple calculation shows that $\cos^2(X)+\sin^2(X)=I$. Can we generalize other trigonometric identities any further?
  • Can this concept be used further to derive some useful results? My senses tell me this should find its place in applied mathematics.

If there's any previous reference (which I think is likely) please inform me.

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Matrix sine and matrix cosine are both $2\pi$-periodic; adding $2\pi\mathbf I$ to your matrix won't change the value of its sine or cosine. As for references, here is one you might want to look at. –  J. M. Aug 25 '12 at 1:39
    
This question is nearly the same as this, and the answers for both seem to give much insights: math.stackexchange.com/questions/80324/… –  progressiveforest Aug 25 '12 at 3:00

2 Answers 2

up vote 4 down vote accepted
  1. Yes, any diagonalizable matrix $T$ whose eigenvalues are integer multiples of $2 \pi i$ has this property, but this does not imply the identity you want in general unless $X$ and $T$ commute. So in particular $T$ can be scalar.

  2. Yes, any trigonometric identity which is a consequence of a polynomial identity between expressions of the form $e^{ix}$ holds also for commuting matrices.

  3. Really the useful operation is the matrix exponential (solving differential equations, relating Lie algebras and Lie groups, etc.) and everything else is just derived from it. But a keyword you might be interested in is "functional calculus."

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in #1, do you know if such $T$ is the only solution? –  progressiveforest Aug 24 '12 at 23:58
    
@progressive: it is. It suffices to verify that if $T$ has nontrivial Jordan blocks then $e^T$ also has nontrivial Jordan blocks. –  Qiaochu Yuan Aug 25 '12 at 0:02
    
Are integer multiples of $2\pi I$ the only periods of this sine function? –  Michael Hardy Aug 25 '12 at 1:47
    
@Michael, yes. If you take the Jordan viewpoint, remember that adding a scalar multiple of the identity shifts the eigenvalues by said scalar multiple; combine this with the knowledge that $\cos\,\mathbf A=\mathbf S(\cos\,\mathbf J)\mathbf S^{-1}$ if $\mathbf A=\mathbf S\mathbf J\mathbf S^{-1}$... –  J. M. Aug 25 '12 at 2:03
    
The fact that those are the only periods seems more interesting than the fact that they are periods. –  Michael Hardy Aug 25 '12 at 13:36

The holomorphic functional calculus provides an algebra homomorphism from functions analytic in a neighbourhood of $\sigma(X)$ (the spectrum of $X$) to the closed subalgebra of the $n \times n$ matrices generated by $X$ (more generally, this is true in any complex Banach algebra). One way to do this is $$f(X) = \frac{1}{2\pi i} \oint_\Gamma f(z) (zI - X)^{-1}\ dz$$ where $\Gamma$ is a contour that surrounds $\sigma(X)$ in an open set where $f$ is analytic. All identities true for ordinary analytic functions carry over to functions of $X$.

In particular, if we choose a locally constant $f$ whose values in a neighbourhood of $\sigma(X)$ are integer multiples of $2 \pi i$, then $f(X)$ satisfies $e^{f(X)} = I$, $\cos(X + f(X)) = \cos(X)$, etc.

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Note the resemblance of the contour integral formulation to the usual Cauchy formula... see this, for instance. –  J. M. Aug 25 '12 at 1:40

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