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