Matrix Calculator

🌐 English
A (3×3)
Rows: 3
Cols: 3
B (3×3)
Rows: 3
Cols: 3
Supports: fractions (1/2), decimals (0.5), constants (pi, e). Empty cells are treated as 0.

Calculation Results

Matrix Arccosine: Definition, Formulas, Examples, and Practice

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.


Definition and Formula

The matrix arccosine of a square matrix \(A\), denoted \(\arccos(A)\), is a matrix function acting as a (local) inverse of the matrix cosine:

$$ \cos(\arccos(A)) = A $$

A convenient relationship is:

$$ \arccos(A) \;=\; \tfrac{\pi}{2}I \;-\; \arcsin(A), $$

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.


Worked Example 1 (Diagonal scalar matrix)

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:

$$ \arccos(B) = \tfrac{\pi}{2}I - B = \begin{pmatrix}\tfrac{\pi}{2} & -1\\[6pt]0 & \tfrac{\pi}{2}\end{pmatrix}. $$

Common Mistakes and Tips

Common mistake Tip / correct approach
Applying scalar arccos element-wise to a non-diagonal matrix Only valid for diagonal matrices or after diagonalization/similar transform that diagonalizes A.
Ignoring branch choices of inverse trig functions Be explicit which branch of \(\arccos\) (principal value) you use; eigenvalues near branch cuts require care.
Naively truncating series for large-norm matrices 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.

Exercise 1 (diagonal small value)

$$ A_1 = \begin{pmatrix}0.1 & 0\\[6pt]0 & 0.1\end{pmatrix} $$
Show Answer

Since \(A_1 = 0.1I\):

$$ \arccos(A_1)=\arccos(0.1)\,I. $$

Numerically \(\arccos(0.1)\approx 1.4706289\) (radians).

Exercise 2 (diagonal larger but inside radius)

$$ A_2 = \begin{pmatrix}0.6 & 0\\[6pt]0 & 0.6\end{pmatrix} $$
Show Answer

Since \(A_2 = 0.6I\):

$$ \arccos(A_2)=\arccos(0.6)\,I. $$

Numerically \(\arccos(0.6)\approx 0.9272952\) (radians).

Exercise 3 (nilpotent)

$$ A_3 = \begin{pmatrix}0 & 1\\[6pt]0 & 0\end{pmatrix} $$
Show Answer

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)

$$ A_4 = \begin{pmatrix}0.2 & 0.3\\[6pt]0 & 0.2\end{pmatrix} $$
Show Answer

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.

First approximation (first term only):

$$ \arccos(A_4) \approx \tfrac{\pi}{2}I - A_4 = \begin{pmatrix}\tfrac{\pi}{2}-0.2 & -0.3\\[6pt]0 & \tfrac{\pi}{2}-0.2\end{pmatrix}. $$

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.


Learn matrix multiplication more easily in just 2 minutes with this free game.

Matrix Multiplication Game