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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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    This question is nearly the same as this, and the answers for both seem to give much insights: http://math.stackexchange.com/questions/80324/sina-where-a-is-a-matrix?rq=12012-08-25

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  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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    @MichaelHardy: as Qiaochu Yuan remarks, $T$ satisfying $e^T=I$ must be diagonalizable with eigenvalues multiples of $2\pi i$. Suppose $T$ is not a multiple of the identity. Let $E = u v^T$ where $u$ is an eigenvector of $T$ for eigenvalue $\alpha$ and $v^T$ is a left eigenvector of $T$ for eigenvalue $\beta \ne \alpha$. We can show by induction that $(E+T)^n = \dfrac{\alpha^n-\beta^n}{\alpha-\beta} E + T^n$ and then $e^{E+T} = I$. On the other hand, $e^E = I+E$. So the only $T$ such that $e^{A+T} = e^A$ for all $n\times n$ matrices $A$ are integer multiples of $2 \pi i I$.2012-08-26
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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](http://www.jstor.org/stable/2306996), for instance.2012-08-25