## Proof exp(A)exp(B) Does Not Equal exp(A+B)

If
$x, \: y$
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
$e^A^B= \left( \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right)$
.
$A+B= \left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right)$
.
To diagonalize
$A+B$
first find the eigenvalues, the solution to
$det((A+B)- \lambda I)=0$
.
\begin{aligned} det((A+B)- \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
$((A+B)- \lambda I) \mathbf{v}= \mathbf{0}$

If
$\lambda =3$
.
$((A+B)- \lambda I) \mathbf{v} = \left( \begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array} \right) \begin{pmatrix}v_1\\v_2\end{pmatrix}= \begin{pmatrix}-v_1+v_2\\v_1-v_2\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix}$
.
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
$D=\left( \begin{array}{cc} 3 & 0 \\ 0 & 1 \end{array} \right)$
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{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^3 & 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^3 & 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^3+e & e^3-e \\ e^3-e & e^3+e \end{array} \right) \end{aligned}
.