\[A\]
(we can only find the sine, or indeed any function of a matrix if it is square), first write down the Mclaurin series for \[sin \: A\]
:\[sin \: A = A- \frac{A^3}{3!}+\frac{A^5}{5!} +...+ (-1)^n \frac{A^{2n-1}}{(2n-11)!}+...\]
.It might be hard to find and add all these powers of
\[A\]
. We can simplify by diagonalising \[A\]
.Example: Let
\[A= \left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right) \]
.To diagonalize
\[A\]
first find the eigenvalues, the solution to \[det(A- \lambda I)=0\]
.\[\begin{equation} \begin{aligned} det(A- \lambda I) &= \left| \begin{array}{cc} 2- \lambda & 1 \\ 1 & 2- \lambda \end{array} \right| \\ &= (2- \lambda)^2-1^2 \\ &= \lambda^2 - 4 \lambda +3 \\ &=(\lambda -3)(\lambda -1) \end{aligned} \end{equation}\]
.Hence
\[\lambda =3, \: \lambda=1 \]
.Now find the eigenvectors. These are the solutions to
\[(A- \lambda I) \mathbf{v}= \mathbf{0}\]
If
\[\lambda =3 \]
.\[(A- \lambda I) \mathbf{v} = \left( \begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array} \right) \begin{pmatrix}v_1\\v_2\end{pmatrix}= \begin{pmatrix}-v_1+v_2\\v_1-v_2\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix}\]
.Hence we can take
\[v_1=v_2=1\]
. The first eigenvector is \[\begin{pmatrix}1\\1\end{pmatrix} \]
.If
\[\lambda =1 \]
.\[(A- \lambda I) \mathbf{v} = \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \begin{pmatrix}v_1\\v_2\end{pmatrix}= \begin{pmatrix}v_1+v_2\\v_1+v_2\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix}\]
.Hence we can take
\[v_1=1, \: v_2=-1\]
. The second eigenvector is \[\begin{pmatrix}1\\-1\end{pmatrix} \]
.Now form the matrix
\[P= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)\]
of eigenvectors and the corresponding diagonal matrix \[D=\left( \begin{array}{cc} 3 & 0 \\ 0 & 1 \end{array} \right)\]
containing the eigenvalues. The relationship between \[A, \: P, \: D\]
is \[D=P^{-1}AP\]
.To find powers of
\[A\]
we can use \[PDP^{-1}=A \rightarrow A^n =\underbrace{(PDP^{-1})...(PDP^{-1})}_{n \: times}=PD^nP^{-1}\]
.\[\begin{equation} \begin{aligned} sin \: A &= PDP^{-1}-\frac{(PDP^{-1})^3}{3!} +...+ (-1)^n\frac{(PDP^{-1})^{2n-1}}{(2n-1)!}+... \\ &= P(D-\frac{D^3}{3!}+\frac{D^5}{5!} +...+ (-1)^n\frac{D^{2n-1}}{(2n-1)!}+...)P^{-1} \\ &= Psin \: D P^{-1} \\ &= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} sin \: 3 & 0 \\ 0 & sin \: 1 \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)^{-1} \\ &= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} sin \: 3 & 0 \\ 0 & sin \: 1 \end{array} \right) \frac{-1}{2} \left( \begin{array}{cc} -1 & -1 \\ -1 & 1 \end{array} \right) \\ &= \frac{1}{2} \left( \begin{array}{cc} sin \: 3+sin \: 1 & sin \: -sin \: 1 \\ sin \: 3-sin \: 1 & sin \: 3+ sin \: 1 \end{array} \right) \end{aligned} \end{equation}\]
.