\[x, \: y\]
are numbers then \[e^x e^y=e^{x+y}\]
. The same relationship does not hold for matrices. That is, if \[A, \: B\]
are matrices then \[e^A e^B \neq e^{A+B}\]
This is a result of the fact that in general matrices
\[A, \: B\]
do not commute, so that \[AB \neq BA\]
.Let
\[A= \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)\]
All powers of
\[A\]
2 and higher are zero, so \[e^A=I+A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)\]
Let
\[B= \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right)\]
All powers of
\[A\]
2 and higher are zero, so \[e^B=I+B=\left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right)\]
Then
\[e^Ae^B= \left( \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right) \]
.\[A+B= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \]
.To diagonalize
\[A+B\]
first find the eigenvalues, the solution to \[det((A+B)- \lambda I)=0\]
.\[\begin{equation} \begin{aligned} det((A+B)- \lambda I) &= \left| \begin{array}{cc} 0- \lambda & 1 \\ 1 & 0- \lambda \end{array} \right| \\ &= (- \lambda)^2-1^2 \\ &= \lambda^2 - 1 \\ &=(\lambda -1)(\lambda +1) \end{aligned} \end{equation}\]
.Hence
\[\lambda =1, \: \lambda= -1 \]
.Now find the eigenvectors. These are the solutions to
\[((A+B)- \lambda I) \mathbf{v}= \mathbf{0}\]
If
\[\lambda =1 \]
.\[((A+B)- \lambda I) \mathbf{v} = \left( \begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array} \right) \begin{pmatrix}v_1\\v_2\end{pmatrix}= \begin{pmatrix}-v_1+v_2\\v_1-v_2\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix}\]
.Hence we can take
\[v_1=v_2=1\]
. The first eigenvector is \[\begin{pmatrix}1\\1\end{pmatrix} \]
.If
\[\lambda =-1 \]
.\[((A+B)- \lambda I) \mathbf{v} = \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \begin{pmatrix}v_1\\v_2\end{pmatrix}= \begin{pmatrix}v_1+v_2\\v_1+v_2\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix}\]
.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 \[D=\left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)\]
containing the eigenvalues. The relationship between \[A+B, \: P, \: D\]
is \[D=P^{-1}(A+B)P\]
.To find powers of
\[A+B\]
we can use \[PDP^{-1}=A+B \rightarrow (A+B)^n =\underbrace{(PDP^{-1})...(PDP^{-1})}_{n \: times}=PD^nP^{-1}\]
.\[\begin{equation} \begin{aligned} e^{A+B} &= I+\frac{PDP^{-1}}{1!}+\frac{(PDP^{-1})^2}{2!} +...+ \frac{(PDP^{-1})^n}{n!}+... \\ &= P(I+\frac{D}{1!}+\frac{D^2}{2!} +...+ \frac{D^n}{n!}+...)P^{-1} \\ &= Pe^DP^{-1} \\ &= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} e^{-1} & 0 \\ 0 & e^1 \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)^{-1} \\ &= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} e^{-1} & 0 \\ 0 & e^1 \end{array} \right) \frac{-1}{2} \left( \begin{array}{cc} -1 & -1 \\ -1 & 1 \end{array} \right) \\ &= \frac{1}{2} \left( \begin{array}{cc} e^{-1}+e & e^{-1}-e \\ e^{-1}-e & e^{-1}+e \end{array} \right) \end{aligned} \end{equation}\]
.