Identity with matrix exponential
Identity with matrix exponential
If $a,b,c$ are real numbers (not all $0$) and denote by $\displaystyle r = \sqrt{a^2+b^2+c^2}$,
Show that: $$\exp{\left( \begin{matrix} 0 & a & c \\ -a & 0 & b \\ -c & -b & 0 \end{matrix} \right)} = \cos{r \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)} + \frac{\sin{r}}{r} \left( \begin{matrix} 0 & a & c \\ -a & 0 & b \\ -c & -b & 0 \end{matrix} \right) + \frac{1 - \cos{r}}{r^2} \left( \begin{matrix} b^{2} & -bc & ab \\ -bc & c^{2} & -ac \\ ab & -ac & a^{2} \end{matrix} \right)$$
Show that: $$\exp{\left( \begin{matrix} 0 & a & c \\ -a & 0 & b \\ -c & -b & 0 \end{matrix} \right)} = \cos{r \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)} + \frac{\sin{r}}{r} \left( \begin{matrix} 0 & a & c \\ -a & 0 & b \\ -c & -b & 0 \end{matrix} \right) + \frac{1 - \cos{r}}{r^2} \left( \begin{matrix} b^{2} & -bc & ab \\ -bc & c^{2} & -ac \\ ab & -ac & a^{2} \end{matrix} \right)$$
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Re: Identity with matrix exponential
Hello r9m.
Let
\(\displaystyle{A=\begin{pmatrix}
0&a &c \\
-a& 0 &b \\
-c&-b & 0
\end{pmatrix}}\).
Then,
\(\displaystyle{\begin{aligned}|A-k\,I_{3}|=0&\iff \begin{vmatrix}
-k&a &c \\
-a& -k &b \\
-c&-b & -k
\end{vmatrix}=0\\&\iff k(k^2+r^2)=0\\&\iff k\in\left\{-i\,r,0,i\,r\right\}\end{aligned}}\)
Using the technology, we find :
\(\displaystyle{A=P\,\Delta\,P^{-1}}\), where :
\(\displaystyle{P=\begin{pmatrix}
b/a& (b\,c+i\,r)/a\,c-i\,r\,b &(b\,c-i\,r\,a)/a\,c+i\,r\,b \\
-c/a& (a^2+b^2)a\,c-i\,r\,b &(a^2+b^2)/a\,c+i\,r\,b \\
1& 1 & 1
\end{pmatrix}}\) .
Now,
\(\displaystyle{e^{A}=P\,e^{\Delta}\,P^{-1}}\), where :
\(\displaystyle{e^{\Delta}=\begin{pmatrix}
1&0 & 0 \\
0&e^{i\,r} &0\\
0& 0 & e^{-i\,r}
\end{pmatrix}}\) .
Check here a relative question.
Let
\(\displaystyle{A=\begin{pmatrix}
0&a &c \\
-a& 0 &b \\
-c&-b & 0
\end{pmatrix}}\).
Then,
\(\displaystyle{\begin{aligned}|A-k\,I_{3}|=0&\iff \begin{vmatrix}
-k&a &c \\
-a& -k &b \\
-c&-b & -k
\end{vmatrix}=0\\&\iff k(k^2+r^2)=0\\&\iff k\in\left\{-i\,r,0,i\,r\right\}\end{aligned}}\)
Using the technology, we find :
\(\displaystyle{A=P\,\Delta\,P^{-1}}\), where :
\(\displaystyle{P=\begin{pmatrix}
b/a& (b\,c+i\,r)/a\,c-i\,r\,b &(b\,c-i\,r\,a)/a\,c+i\,r\,b \\
-c/a& (a^2+b^2)a\,c-i\,r\,b &(a^2+b^2)/a\,c+i\,r\,b \\
1& 1 & 1
\end{pmatrix}}\) .
Now,
\(\displaystyle{e^{A}=P\,e^{\Delta}\,P^{-1}}\), where :
\(\displaystyle{e^{\Delta}=\begin{pmatrix}
1&0 & 0 \\
0&e^{i\,r} &0\\
0& 0 & e^{-i\,r}
\end{pmatrix}}\) .
Check here a relative question.
Re: Identity with matrix exponential
Hi Papapetros Vaggelis, Nice solution! Looking at the Eigenvalues indeed is the simplest approach!
You derived the characteristic equation to be: $k(k^2+r^2) = 0$, so by Cayley-Hamilton Theorem we have the identity $A^3+r^2A = 0$.
Thus in general we have: $$\begin{align*}A^{2n}=(-1)^{n-1}r^{2n-2}A^2 \\A^{2n-1}=(-1)^{n-1}r^{2n-2}A\end{align*}$$
Thus, $$\begin{align*}e^{A} = I_3 +\sum\limits_{n=1}^{\infty} \frac{A^n}{n!} &= I_3 + \sum\limits_{n=1}^{\infty} \frac{A^{2n}}{(2n)!}+\sum\limits_{n=1}^{\infty} \frac{A^{2n-1}}{(2n-1)!}\\&= I_3 +\left(\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}r^{2n-2}}{(2n)!}\right)A^2 + \left(\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}r^{2n-2}}{(2n-1)!}\right)A\\&= I_3 + \frac{\sin r}{r}A+\frac{1-\cos r}{r^2}A^2\end{align*}$$
You derived the characteristic equation to be: $k(k^2+r^2) = 0$, so by Cayley-Hamilton Theorem we have the identity $A^3+r^2A = 0$.
Thus in general we have: $$\begin{align*}A^{2n}=(-1)^{n-1}r^{2n-2}A^2 \\A^{2n-1}=(-1)^{n-1}r^{2n-2}A\end{align*}$$
Thus, $$\begin{align*}e^{A} = I_3 +\sum\limits_{n=1}^{\infty} \frac{A^n}{n!} &= I_3 + \sum\limits_{n=1}^{\infty} \frac{A^{2n}}{(2n)!}+\sum\limits_{n=1}^{\infty} \frac{A^{2n-1}}{(2n-1)!}\\&= I_3 +\left(\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}r^{2n-2}}{(2n)!}\right)A^2 + \left(\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}r^{2n-2}}{(2n-1)!}\right)A\\&= I_3 + \frac{\sin r}{r}A+\frac{1-\cos r}{r^2}A^2\end{align*}$$
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