Sine of a Matrix

The sine of a matrix $ A $ is defined using the Taylor series expansion for sine: $$ \sin(A) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} A^{2k+1} $$

To compute the sine of a matrix, we evaluate this infinite series by calculating the powers of $ A $ and applying the corresponding coefficients.

This series converges for any matrix $ A $.

Note. The sine of a matrix has various applications, especially in physics and engineering, where it helps model dynamic systems and solve matrix differential equations.

A Practical Example

Consider the matrix $ A $:

$$ A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$

This matrix represents a rotation and has a straightforward structure, making the calculations relatively simple.

Now, we can compute $ \sin(A) $ using the first few terms of the Taylor series:

$$ \sin(A) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} A^{2k+1} $$

The series for $ \sin(A) $ expands as:

$$ \sin(A) = A - \frac{A^3}{3!} + \frac{A^5}{5!} - \dots $$

Observe that the square of $ A $ is:

$$ A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I $$

where $ I $ is the identity matrix.

Note. To compute the power of a matrix, we multiply $ A $ by itself, rather than raising each element to the power. To calculate $ A^2 $, we proceed with matrix multiplication as follows:
$$ A^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$ $$ = \begin{pmatrix} 0 \cdot 0 + 1 \cdot (-1) & 0 \cdot 1 + 1 \cdot 0 \\ -1 \cdot 0 + 0 \cdot (-1) & -1 \cdot 1 + 0 \cdot 0 \end{pmatrix} $$ $$ = \begin{pmatrix} 0 - 1 & 0 + 0 \\ 0 + 0 & -1 + 0 \end{pmatrix} $$ So, the result is: $$ A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} $$ This outcome matches $-I$, where $ I $ is the identity matrix. $$ A^2 = - I $$

Knowing $ A^2 = -I $, it follows that $ A^3 $ can be found as:

$$ A^3 = A \cdot A^2 = A \cdot (-I) = -A $$

Similarly, we find that $ A^4 = I $:

$$ A^4 = A^2 \cdot A^2 = (-I) \cdot (-I) = I $$

This means $ A^5 = A $, since multiplying any matrix $ A $ by the identity matrix $ I $ yields $ A $:

$$ A^5 = A \cdot A^4 = A \cdot I = A $$

Using $ A^3 = -A $ and $ A^5 = A $, we can rewrite the series as:

$$ \sin(A) = A - \frac{A^3}{3!} + \frac{A^5}{5!} - \dots $$

$$ \sin(A) = A - \frac{(-A)}{3!} + \frac{A}{5!} - \dots $$

Summing the first few terms, which repeat cyclically, gives a good approximation for $ \sin(A) $:

$$ \sin(A) \approx A \left(1 - \frac{1}{3!} + \frac{1}{5!} - \dots \right) $$

This approach converges quickly, offering a practical approximation of $ \sin(A) $.

And so forth.