10.2: La Matriz Exponencial como Límite de Poderes
- Page ID
- 113046
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Puede recordar de Cálculo que para cualquier número aa y tt uno puede lograr\(e^{at}\) a través de
\[e^{at} = \lim_{k \rightarrow \infty} (1+\frac{at}{k})^k \nonumber\]
Por lo tanto, la definición de matriz natural es
\[e^{At} = \lim_{k \rightarrow \infty} (I+\frac{At}{k})^k \nonumber\]
donde\(I\) es la matriz de identidad n-por-n.
El caso más fácil es el caso diagonal, por ejemplo,
\[A = \begin{pmatrix} {1}&{0}\\ {0}&{2} \end{pmatrix} \nonumber\]
para entonces
\[(I+\frac{At}{k})^k = \begin{pmatrix} {(1+\frac{t}{k})^k}&{0}\\ {0}&{(1+\frac{2t}{k})^k} \end{pmatrix} \nonumber\]
y así
\[e^{At} = \begin{pmatrix} {e^t}&{0}\\ {0}&{e^{2t}} \end{pmatrix} \nonumber\]
Tenga en cuenta que este NO es el exponencial de cada elemento de\(A\).
Como ejemplo concreto supongamos
\[A = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} \nonumber\]
Desde
\[I+At = \begin{pmatrix} {1}&{t}\\ {-t}&{1} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^2 = \begin{pmatrix} {1}&{\frac{t}{2}}\\ {\frac{-t}{2}}&{1} \end{pmatrix} \begin{pmatrix} {1}&{\frac{t}{2}}\\ {\frac{-t}{2}}&{1} \end{pmatrix} = \begin{pmatrix} {1-\frac{t^2}{4}}&{t}\\ {-t}&{1-\frac{t^2}{4}} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^3 = \begin{pmatrix} {1-\frac{t^2}{3}}&{t-\frac{t^3}{27}}\\ {-t+\frac{t^3}{27}}&{1-\frac{t^2}{3}} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^4 = \begin{pmatrix} {-\frac{3t^2}{8}+\frac{t^4}{256}+1}&{t-\frac{t^3}{16}}\\ {-t+\frac{t^3}{16}}&{-\frac{3t^2}{8}+\frac{t^4}{256}+1} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^5 = \begin{pmatrix} {-\frac{2t^2}{5}+\frac{t^4}{125}+1}&{t-\frac{2t^3}{25}+\frac{t^5}{3125}}\\ {-t+\frac{2t^3}{25}-\frac{t^5}{3125}}&{-\frac{2t^2}{5}+\frac{t^4}{125}+1} \end{pmatrix} \nonumber\]
Discernimos un patrón: los elementos diagonales son iguales polinomios pares mientras que los elementos diagonales son iguales pero opuestos polinomios impares. El grado del polinomio crecerá con kk y en el límite 'reconocemos'
\[e^{At} = \begin{pmatrix} {\cos(t)}&{-\sin(t)}\\ {\sin(t)}&{\cos(t)} \end{pmatrix} \nonumber\]
Si
\[A = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]
entonces
\[(I+\frac{At}{k})^k = \begin{pmatrix} {1}&{t}\\ {0}&{1} \end{pmatrix} \nonumber\]
para cada valor de\(k\) y así
\[e^{At} = \begin{pmatrix} {1}&{t}\\ {0}&{1} \end{pmatrix} \nonumber\]