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Arcocoseno matricial: definición, fórmulas, ejemplos y práctica
This article explains the matrix arccosine function \(\arccos(A)\), shows
definitions and practical formulas, includes two worked examples, lists common pitfalls, and
provides four practice problems with collapsible answers. It is intended as companion
content for a Matrix Arccosine Calculator.
Definición y fórmula
The matrix arccosine of a square matrix \(A\), denoted \(\arccos(A)\), is a matrix function acting as a (local) inverse of the matrix cosine:
$$ \cos(\arccos(A)) = A $$Una relación útil es:
$$ \arccos(A) \;=\; \tfrac{\pi}{2}I \;-\; \arcsin(A), $$so one can compute \(\arccos(A)\) by computing \(\arcsin(A)\) (for a chosen branch) and subtracting from \(\tfrac{\pi}{2}I\). The matrix arcsine admits the Taylor series (for small \(\|A\|\)):
$$ \arcsin(A) = A + \tfrac{1}{6}A^{3} + \tfrac{3}{40}A^{5} + \tfrac{5}{112}A^{7} + \cdots, $$thus
$$ \arccos(A) = \tfrac{\pi}{2}I - \Big(A + \tfrac{1}{6}A^{3} + \tfrac{3}{40}A^{5} + \cdots\Big). $$In practice, numerical and stable methods use diagonalization, Jordan/Schur decompositions, or tailored matrix-function routines rather than naive series truncation.
Ejemplo resuelto 1 (matriz escalar diagonal)
Let
$$ A = \begin{pmatrix} 0.2 & 0 \\[6pt] 0 & 0.2 \end{pmatrix}. $$Because \(A = 0.2 I\), apply scalar arccos element-wise:
$$ \arccos(A) = \arccos(0.2)\, I. $$Numerically, \(\arccos(0.2)\approx 1.369438\) (radians), so
$$ \arccos(A) \approx 1.369438 \begin{pmatrix}1 & 0\\[6pt]0 & 1\end{pmatrix}. $$Worked Example 2 (Nilpotent matrix — series truncation)
Let
$$ B = \begin{pmatrix}0 & 1\\[6pt]0 & 0\end{pmatrix}. $$Note \(B^2 = 0\). Using relation \(\arccos(B)=\tfrac{\pi}{2}I-\arcsin(B)\) and the fact \(\arcsin(B)=B\) (series truncates because \(B^3=0\)), we get:
$$ \arccos(B) = \tfrac{\pi}{2}I - B = \begin{pmatrix}\tfrac{\pi}{2} & -1\\[6pt]0 & \tfrac{\pi}{2}\end{pmatrix}. $$Errores comunes y consejos
| Common mistake | Tip / correct approach |
|---|---|
| Aplicar el arcocoseno escalar elemento a elemento a una matriz no diagonal | Only valid for diagonal matrices or after diagonalization/similar transform that diagonalizes A. |
| Ignorar la elección de ramas de las funciones trigonométricas inversas | Be explicit which branch of \(\arccos\) (principal value) you use; eigenvalues near branch cuts require care. |
| Truncar ingenuamente la serie para matrices de norma grande | Use Schur or diagonalization methods for stability or scale-and-square approaches. |
| Assuming \(\arccos(A)\) always exists | Matrix function inverse exists only when matrix lies in the range of \(\cos\) on chosen domain; check spectrum and convergence. |
Practice Problems (Compute \(\arccos(A)\))
The exercises below are suitable for hand computation when possible or for use with your Matrix Arccosine Calculator. Matrices are visible; click "Show Answer" to reveal solutions.
Ejercicio 1 (diagonal, valor pequeño)
$$ A_1 = \begin{pmatrix}0.1 & 0\\[6pt]0 & 0.1\end{pmatrix} $$Mostrar respuesta
Since \(A_1 = 0.1I\):
$$ \arccos(A_1)=\arccos(0.1)\,I. $$Numerically \(\arccos(0.1)\approx 1.4706289\) (radians).
Ejercicio 2 (diagonal, mayor pero dentro del radio)
$$ A_2 = \begin{pmatrix}0.6 & 0\\[6pt]0 & 0.6\end{pmatrix} $$Mostrar respuesta
Since \(A_2 = 0.6I\):
$$ \arccos(A_2)=\arccos(0.6)\,I. $$Numerically \(\arccos(0.6)\approx 0.9272952\) (radians).
Ejercicio 3 (nilpotente)
$$ A_3 = \begin{pmatrix}0 & 1\\[6pt]0 & 0\end{pmatrix} $$Mostrar respuesta
Because \(A_3^2=0\), the series for \(\arcsin\) truncates at the linear term: \(\arcsin(A_3)=A_3\). Therefore
$$ \arccos(A_3)=\tfrac{\pi}{2}I - A_3 = \begin{pmatrix}\tfrac{\pi}{2} & -1\\[6pt]0 & \tfrac{\pi}{2}\end{pmatrix}. $$Exercise 4 (non-diagonal, small norm — series method)
$$ A_4 = \begin{pmatrix}0.2 & 0.3\\[6pt]0 & 0.2\end{pmatrix} $$Mostrar respuesta
Approach 1 (series): compute \(\arcsin(A_4)=A_4 + \tfrac{1}{6}A_4^3 + \cdots\) and then \(\arccos(A_4)=\tfrac{\pi}{2}I-\arcsin(A_4)\). Because \(\|A_4\|<1\) the series converges.
Primera aproximación (solo el primer término):
$$ \arccos(A_4) \approx \tfrac{\pi}{2}I - A_4 = \begin{pmatrix}\tfrac{\pi}{2}-0.2 & -0.3\\[6pt]0 & \tfrac{\pi}{2}-0.2\end{pmatrix}. $$For higher accuracy compute \(A_4^3\) and add the \(\tfrac{1}{6}A_4^3\) correction to \(\arcsin(A_4)\) before subtracting from \(\tfrac{\pi}{2}I\), or use a Schur-based matrix-function routine in the calculator.
El arcocoseno matricial también se conecta bien con el rango de una matriz, la SVD y la RREF.