## Arcsin of a Matrix

We could find the inverse sine of a matrix using Mclaurin series, but here is an alternative. We find
$f(A)=sin^{-1}(\frac{A}{4})$
.
Let
$m( \lambda)$
be the minimum polynomial of the matrix
$A$
. Suppose we can find a polynomial
$r(\lambda)$
which has degree less than the degree of the minimum polynomial satisfying
$f(\lambda)=r(\lambda)m \: f'(\lambda)=r'(\lambda)$
for each eigenvalue
$\lambda$
of
$A$
. Then
$r(A)=f(A)$
, where
$f(A)$
is a matrix polynomial. NOW, to evaluate
$sin^{-1} (\frac{A}{4})$
, first find the minimum polynomial of
$A$
. The characteristic polynomial of
$A$
is
$det(A- \lambda I)= \left| \begin{array}{cc} 0- \lambda & -1 \\ 4 & 4- \lambda \end{array} \right| = (2- \lambda)^2$
and this is the minimum polynomial since
$A- 2I$
is not equal to the zero matrix. The polynomial
$r(\lambda)$
must be of degree less than the degree of the minimum polynomial less than the or r (2) sin that is, 01 and The polynomial r (X) must be of degree degree of the minimum polynomial, so let
$r(\lambda) = \alpha + \beta \lambda$

We have
$r(\lambda)=f(\lambda)$
so
$r(2)= f(2)=sin^{-1} (\frac{2}{4})$

Also
$f'(\lambda) = f'(\lambda)$
so
$r'(2) = \beta =\frac{d}{d \lambda} (sin^{-1} (\frac{\lambda}{4}) |_{\lambda =2}= \frac{\sqrt{3}}{6}$

Then
$\alpha + 2 \beta = sin^{-1}( \frac{1}{2}) = \frac{\pi}{6}$

Then
$\alpha = \frac{\pi}{6}- 2 \beta = \frac{\pi}{6}- 2 (\frac{\sqrt{3}}{6} =\frac{1}{6}( \pi - 2 \sqrt{3})$

Then
$r(\lambda)=\frac{1}{6}( \pi - 2 \sqrt{3} +{\sqrt{3}} \lambda)$

Hence
\begin{aligned} r(A) &= f(A) \\ &= sin^{-1} (\frac{A}{4}) \\ &= \frac{1}{6}( (\pi - 2 \sqrt{3})I +{\sqrt{3}} A)\\ &= \frac{1}{6} (( \pi - 2 \sqrt{3}) \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) + \sqrt{3} \left( \begin{array}{cc} 0 & -1 \\ 4 & 4 \end{array} \right)) \\ &= \frac{1}{6} \left( \begin{array}{cc} \pi - 2 \sqrt{3} & - \sqrt{3} \\ 4 \sqrt{3} & \pi + 2 \sqrt{3} \end{array} \right) \end{aligned}