## The Sine of a Matrix

To find the sine of a square matrix
$A$
(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{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}
.
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{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}
.