§8. Matrix Differentials and Matrix Lie Groups

A matrix-valued function, or simply a matrix function, can either be interpreted as a function taking $m\times n$ matrices as values or as an $m\times n$ matrix with (scalar-valued) functions as entires, say $\Phi=[f^j_k]$, and its differential is obtained by simply taking the differentials of its entries: $d\Phi=[df^j_k]$; here we use the superscript $j$ and the subscript $k$ to indicate the $j$th row and $k$th column.

Example 8.1. Find $d\Phi$ at $t=0$ for the matrix function $$\Phi=\left[\begin{array}{rr} \cos t & -\sin t \\ \sin t & \cos t \end{array}\right]$$.

Solution. $d\Phi=\left[\begin{array}{rr} d\cos t & -d\sin t \\ d\sin t & d\cos t \end{a... ...ray}{rr} -\sin dt & -\cos t dt \\ \cos t dt & -\sin t dt \end{array}\right]$. At $t=0$, we have

$$d\Phi=\left[\begin{array}{rr} 0 & -dt \\ dt & 0 \end{array... ...ft[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right] dt,$$

which is a skew symmetric matrix times $dt$.

Suppose that we have anothe matrix function, say $\Psi=[g^j_k]$, which is $n\times p$. Then $\Phi\Psi$ is an $m\times p$ matrix function with its $(j,k)$ entry is given by $h^j_k={\sum}_{i=1}^n f^j_ig^i_k$. Notice that

$$dh^j_k={\sum}_{i=1}^n\ d(f^j_ig^i_k)={\sum}_{i=1}^n\ f_j^idg_i^k+ {\sum}_{i=1}^n\ (d f^j_i)g^i_j.$$

This verifies the following

Rule No. M1 (Product rule for matrix function). $d(\Psi\Phi)=\Psi.\ d\Phi+d\Psi.\ \Phi$.

Caveat: the order of the product $d\Psi.\ \Phi$ CANNOT be reversed. For scalar functions $f$ and $g$ we do have $(df)g=gdf$. But this is no longer valid for matrix functions.

Example 8.2. Let $\Phi$ be an invertible $n\times n$ matrix function. (``Invertible'' here means that there is a matrix function $\Phi^{-1}$, called the inverse of $\Phi$ such that $\Phi^{-1}\Phi=\Phi\Phi^{-1}=I$, where $I$ is the identity matrix.) Verify that

$$d(\Phi^{-1})=-\Phi^{-1}.\ d\Phi.\ \Phi^{-1}. \ \ \ \ \ \ \ (8.1)$$

(Caveat: the last expression cannot be rewritten as $-\Phi^{-2}d\Phi$.)

Solution. Differntiate both sides of $\Phi^{-1}\Phi=I$, the RHS becomes $O$ because $I$ is a constant matrix and the LHS is $d(\Phi^{-1}\Phi)=d\Phi^{-1}.\ \Phi+\Phi^{-1}.\ d\Phi$. Thus $d\Phi^{-1}.\ \Phi=-\Phi^{-1}.\ d\Phi$. Multiply both sides on the right by $\Phi^{-1}$, we have $d\Phi^{-1}=-\Phi^{-1}.\ d\Phi.\ \Phi^{-1}$.

A particular case of Rule No. M1 is $d(\Phi {\bf x})=\Phi d{\bf x}+d\Phi. {\bf x}$, where $\bf x$ is a column vector. This is because a column vector here is nothing but an $n\times 1$ matrix. We will identify a column vector with a usual vector with components displayed in a row by means of the following convention: any row of objects surround by the round brackets ``('' and ``)'' is the same as the column of same objects, following the same order, embraced by square brackets ``['' and ``]''. For example, if $A=\left[\begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array}\right]$ and ${\bf x}=({\rm cats},\ {\rm dogs})$, then

$$A{\bf x}=\left[\begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array... ...{\rm dogs} \\ 4\ {\rm cats}+5\ {\rm dogs} \end{array}\right]$$

and hence $A{\bf x}=(2\ {\rm cats}+3\ {\rm dogs},\ 4\ {\rm cats} +5\ {\rm dogs})$. (No more silly examples of this sort!)

We say that a collection $G$ of $n\times n$ complex matrices form a matrix group if it satisfies the following conditions:

(MG1) The identity matrix $I$ belongs to $G$.

(MG2) If $S$ and $T$ are in $G$, then so is thier product $ST$.

(MG3) If $S$ is in $G$, then $S$ is invertible and its inverse $S^{-1}$ is also in $G$.

Our purpose of introducing matrix groups is to let the reader have a taste of Lie groups, which is an important mathematical subject. (``Lie'' here is pronounced as ``Lee'').

Example 8.3. Show that all invertible real $n\times n$ matrices form a matrix group. This matrix group is denoted by ${\rm GL}(n, {\bf R})$, called the general linear group.

Solution. We have to check conditions (MG1) to (MG3) for ${\rm GL}(n)$. Clearly the identity matrix is invertible. If $S$ and $T$ are invertible $n\times n$ matrices, then so is $ST$ with $(ST)^{-1}=T^{-1}S^{-1}$. Indeed $(ST)(T^{-1}S^{-1})=S(TT^{-1})S^{-1}=SIS^{-1}=SS^{-1}=S$ and similarly $(T^{-1}S^{-1})(ST)=I$. It remains to show MG3). This follows from the fact that if $S$ is invertible, then so is $S^{-1}$ and the inverse of $S^{-1}$ is just $S$ itself. (As you can see, the important thing here is to know what you must check. The actual checking is easy and rather insipid. This situation is a bit like dealing with bureaucrats - it is important to know the right person to talk to, but you do not expect the conversation to be interesting.)

Example 8.4. Show that the set $O(n)$ of all real $n\times n$ matrices $R$ satisfying $RR^\top=R^\top R=I$ is a matrix group. This matrix group is called the orthogonal group. (Recall that $R^\top$ is the transpose of $R$. A matrix $R$ satisfying the condition $RR^{\top}=R^\top R=I$ (that is, $R^{-1}=R^\top$) is called an orthogonal matrix.)

Solution. We have to check that the set $O(n)$ satisfies conditions (MG1) to (MG3). Since $I^{-1}=I=I^\top$, it follows that the identity matrix $I$ is orthogonal. Let $R$ and $S$ be orthogonal matrices. Then $RR^\top=R^\top R=I$ and $SS^\top=S^\top S=I$. Hence $(RS)(RS)^\top=(RS)(S^\top R^\top)=R(SS^\top)R^\top=RR^\top=I$ and similarly $(RS)^\top(RS)=I$; (here we have used the well known identity $(RS)^\top=S^\top R^\top$ in matrix theory). Finally, suppose that $R$ is an orthogonal matrix. Then $R^{-1}=R^\top$ and hence $R^{-1}(R^{-1})^{\top}=R^{-1}R^{\top\top}=R^{-1}R=I$ and simialrly $(R^{-1})^\top R^{-1}=I$, showing that $R^{-1}$ is also orthogonal.

Now we introduce an important concept: the Lie algebra ${\rm L}G$ of a (closed) matrix group $G$. An element in ${\rm L}G$ is an $n\times n$ matrix produced in the following way. Take any parametric curve $\Phi(t)$ in $G$ such that $\Phi(0)=I$. Differentiate $\Phi$ and evaluate the derivative $\Phi^\prime$ at $0$. The resulting matrix $A=\Phi^\prime(0)$ is an element in ${\rm L}G$. For example,

$$R(t)=\left[\begin{array}{rr} \cos t & -\sin t \\ \sin t & ... ...ight], \ \ \ \ -\infty<t<\infty, \ \ \ \ \ \ \ \ \ \ \ (8.1)$$

is in $O(2)$ with $R(0)=I$; (the reader should check $R(t)^\top R(t)=R(t)R(t)^\top=I$). Hence

$$R^{\prime}(0)=\frac{d\ }{dt}\ R(t)\ \bigg\vert _{t=0}=\left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right]$$

belongs to the Lie algebra ${\bf o}(2)={\rm L}O(2)$. Now let us return to the general discussion. Take arbitrary $A$ and $B$ in ${\rm L}G$. We claim: $A+B$ is also in ${\rm L}G$. Why? Well, by assumption, we have $\Phi^\prime(0)=A$ and $\Psi^\prime(0)=B$ for some parametric curves $\Phi$ and $\Psi$ in $G$ with $\Phi(0)=\Psi(0)=I$. Then $\Theta(t)\equiv\Phi(t)\Psi(t)$ is also a parametric curve in $G$ with $\Theta(0)=\Phi(0)\Psi(0)=I$. So $\Theta^\prime(0)$ is in ${\rm L}G$. The product rule gives

$$\Theta^\prime(0) =\Phi(0)\Psi^\prime(0)+\Phi^\prime(0)\Psi(0)=I.B+A.I=A+B.$$

Hence $A+B$ is in ${\rm L}G$. Next we claim: if $A$ is in ${\rm L}G$, say $A=\Phi^\prime(0)$ for a parametric curve $\Phi(t)$ with $\Phi(0)=I$, and if $\lambda$ is a real number, then $\lambda A$ is also in ${\rm L}G$. Indeed, Consider the new parametric curve $\Psi(t)=\Phi(\lambda t)$, which is also lying in $G$. Clearly $\Psi(0)=I$ and, by the chain rule, $\Psi^\prime(t)=\lambda\Phi^\prime(\lambda t)$ and consequently $\lambda A=\lambda \Phi^\prime(0)=\Psi^\prime(0)\in{\rm L}G$. We make our third claim, which is very interesting: if $S$ is in $G$ and if $A$ is in ${\rm L}G$, then $S^{-1}AS$ is also in ${\rm L}G$. Indeed, if $\Phi(t)$ is a parametric curve in $G$ with $\Phi(0)=I$ and $\Phi^\prime(0)=A$, then $\Theta(t)=S^{-1}\Phi(t)S$ is also a parametric curve in $G$ with $\Theta(0)=S^{-1}\Phi(0)S=S^{-1}IS=I$ and $\Theta^\prime(0)=S^{-1}\Phi(0)S=S^{-1}AS$, showing that $S^{-1}AS$ also belongs to ${\rm L}G$. Our fourth and final claim: if $A$ and $B$ are in ${\rm L}G$, then $AB-BA$ is also in ${\rm L}G$. Indeed, take a parametric curve $\Psi(t)$ in $G$ such that $\Psi(0)=I$ and $\Psi^\prime(0)=B$. By our third claim, we know that $C(t)=\Psi(t)^{-1}A\Psi(t)$ is a parametric curve in ${\rm L}G$. Now our first and second claims say that ${\rm L}G$ is a linear space of $n\times n$ matrices. Hence the derivative $C^\prime(t)$ of $C(t)$, a curve in ${\rm L}G$, is also in ${\rm L}G$. In particular, $C^\prime(0)$ is in ${\rm L}G$. What is $C^\prime(0)$? Well,

$$C^{\prime}=(\Phi^{-1}A\Phi)^\prime=(\Phi^{-1})^{\prime}A\Phi+... ...ime=-\Phi^{-1}\Phi^\prime \Phi^{-1}A\Phi+\Phi^{-1}A\Phi^\prime;$$

(here we have used $(\Phi^{-1})^\prime=-\Phi^{-1}\Phi^\prime\Phi$, which follows from Example 8.2.) From $\Phi(0)=I$ and $\Phi^\prime(0)=A$, we obtain $C^\prime(0)=-BA+AB=AB-BA$. The expression $AB-BA$ is called the Lie bracket or Lie product of $A$ and $B$, and is denoted by $[A,B]$. We call a set $L$ of $n\times n$ matrices a (matrix) Lie algebra if, for all $A$ and $B$ in $L$ and for all real numbers $\lambda$ and $\mu$, both $\lambda A+\mu B$ and $[A,B]$ are in $L$. We have arrived at:

Fact. If $G$ is a (closed) matrix group, then ${\rm L}G$ is a Lie algebra, and we have

$$S\in G, \ \ {\rm and} \ \ A\in {\rm L}G \ \ \ {\rm imply} \ \ \ S^{-1}AS\in G.$$

Usually we write ${\rm Ad}(S)$ for the linear operator on ${\rm L}G$ sending $A$ to $S^{-1}AS$. Then the map $S\mapsto {\rm Ad}(S)$ is a representation of $G$, called the adjoint representation of $G$.

Example 8.5. A matrix Lie group $G$ is said to be commutative or abelian if $ST=TS$ for all $S$ and $T$ in $G$. Show that, if $G$ is an abelian matrix group, then its ajoint representaion is trivial in the sense that ${\rm Ad}(S)$ is the identity mapping (that is ${\rm Ad}(S)A=A$ for all $A\in {\rm L}G$) for all $S\in G$ and its Lie algebra ${\rm L}G$ is also abelian in the sense that $[A,B]=O$ for all $A$ and $B$ in ${\rm L}G$.

Solution: Let $A\in {\rm L}G$ and $S\in G$. Then there is a parametric curve $\Phi(t)$ in $G$ such that $\Phi(0)=I$ and $\Phi^\prime(0)=A$. Since $G$ is commutative, we have $S\Phi(t)=\Phi(t)S$. Differentiating the last identity with respect to $t$ and then evaluating at $t=0$, we arrive at $SA=AS$. Thus ${\rm Ad}(S)A\equiv S^{-1}AS=A$. Next, let $B$ be another element of the Lie algebra ${\rm L}G$ and $\Psi(t)$ be a corresponding parametric curve in $G$ such that $\Psi(0)=I$ and $\Psi^\prime(0)=B$. Then we have $\Psi(t)A=A\Psi(t)$. Differentiating the last identity and setting $t=0$, we have $BA=AB$. Therefore $[A,B]=AB-BA=O$.

In our definition of the Lie algebra ${\rm L}G$ of a matrix group, we say that $A$ is in it if and only if there exists a curve $\Phi(t)$ in $G$ such that $\Phi(0)=0$ and $\Phi^\prime(0)=A$. A natural question is: is there a good choice of $\Phi(t)$ for us to check this ? The answer is YES. It is $\Phi(t)=e^{tA}$. It turns out that $A$ is in ${\rm L}G$ if and only if $e^{tA}$ is in $G$ for all $t$. But what is the meaning of $e^{tA}$ ? To define the matrix exponential $e^A$, we mimick the scalar case $e^a$, where $a$ is a real number. You probably have never worried about how to define $e^a$ rigorously. But now we have to. Recall from elementary calculus that $y=e^{at}$ ($e^a$ is just $e^{at}$ when $t=1$) is the solution to the initial value problem $y^\prime=ay$, $y(0)=1$. So, if we can establish the existence and uniqueness of solution to this initial value problem, then we can define $e^{at}$ to be the unique its solution. We adopt the same tactic in our treatment for matrix exponentials. Take any $n\times n$ (constant) matrix $A$. Consider the matrix differential equation $\Phi^\prime=A\Phi$, where the unknown $\Phi$ is an $n\times n$ matrix function, with the initial condition $\Phi(0)=I$. Thus we are considering the following initial value problem for a matrix equation:

$$\Phi^\prime=A\Phi, \ \ \ \ \Phi(0)=I. \ \ \ \ \ \ \ \ \ \ \ (8.2)$$

If $\Phi(t)$ is a solution, then $\Phi(t)=\Phi(0)+\int_0^t\Phi^{\prime}(s)ds=I+\int_0^tA\Phi(s)ds =I+A\int_0^t\Phi(s)ds$. Thus we can convert the initial value problem (8.2) to a Volterra type integral equation:

$$\Phi(t)=I+A\int_0^t\Phi(s)ds. \ \ \ \ \ \ \ \ \ \ \ (8.3)$$

There are two advantages of (8.3) over (8.2). First, the initial condition $\Phi(0)=I$ is automatically accounted for in (8.3). Second, we can apply the Picard's iteration method - the philosophy behind this is that in general differentiation drives functions wild, and integration tames them. Starting with any matrix function $\Phi_0(t)$, we produce a sequence $\Phi_n(t)$ by repeated substitution: $\Phi_{n+1}(t)=I+A\int_0^t\Phi_n(s)ds$. The limit $\Phi(t)=\lim_{n\to\infty}\Phi_n(t)$ is a solution to (8.3) and hence to (8.2). This method will be justified in the last section of the present chapter. Here let us just apply it to see what it will give. Let us begin with $\Phi_0(t)=I$. Then $\Phi_1(t)=I+A\int_0^t\Phi_0(s)ds=I+At$. Next,

$$\begin{eqnarray*} \Phi_2(t) &=&I+A\int_0^t\Phi(s)ds=I+A\int_0^t(I+As)ds \\ &=&I+A(It+At^2/2) =I+At+A^2t^2/2. \end{eqnarray*}$$

Continue in the manner, in general, we have $\Phi_n(t)=I+At+A^2t^2/2!+\cdots+A^nt^n/n!$ (which can be verified by the method of induction). Letting $n\to\infty$, we obtain the ``series solution'' to (8.2), which gives the definition of the matrix exponential $e^{At}$:

$$e^{At}=\lim_{n\to \infty}\Phi_n(t) =I+At+\frac{A^2t}{ 2!}+\c... ...s\ \equiv{\sum}_{n=0}^\infty\ \frac{A^nt^n}{ n!}. \ \ \ (8.4)$$

A general method for computing matrix exponential is given in Appendix A. Here we only consider simple examples for which this method is not needed.

Example 8.6. Find $e^{At}$ for $A=\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]$.

Solution: A simple computation shows that $A^2=-I$, $A^3=-A$ and $A^4=I$. Thus the sequence $A^n$ ( $n=0, 1,2,\ldots$ ) repeats the pattern of $I, \ A,\ -I,\ -A$. So

$$\begin{eqnarray*}\Phi(t) &\equiv& e^{At}=I+At-I\frac{t^2}{2!}-A\frac{t^3}{ 3!}+... ...!}+\cdots\ \right)I+ \left(t-\frac{t^3}{ 3!}+\cdots\ \right)A \end{eqnarray*}$$

Recalling the series expansion for $\cos t$ and $\sin t$, we obtain $\Phi(t)=I\,\cos t+A\,\sin t$, which is the rotation matrix given by (8.1).

One of our purposes of introducing the matrix exponential is to present the following

Fact. $A$ is in ${\rm L}G$ if and only if $e^{At}$ is in $G$ for all $t$.

Suppose that $e^{At}$ is in $G$ for all $t$. Then $\Phi(t)=e^{At}$ is a parametric curve in $G$ with $\Phi(0)=I$ and hence $\Phi^\prime(0)=A$ is in ${\rm L}G$. This proves the ``if'' part. The ``only if'' part is much harder to prove. We only give a very brief sketch here. Suppose that $A\in {\rm L}G$. Then there is a smooth parametric curve $\Phi(t)$ such that $\Phi(0)=I$ and $\Phi^\prime(0)=A$. Thus, near $t=0$, we have $\Phi(t)=I+(A+\alpha(t))t$, where $\alpha(t)$ shrinks to $0$ as $t\to 0$. Now fix $t$. For large $n$,

$$\Phi(t/n)^n=(I+(A+\alpha(t/n))t/n)^n\approx (I+At/n)^n \to e^{At} \ \ {\rm as}\ n\to \infty.$$

Since $\Phi(t/n)^n$ is in $G$ for each $n$ and since $G$ is closed, we have $e^{At}$ in $G$ for all $t$.

Example 8.7. Determine the Lie algebra ${\bf o}(n)$ of the orthogonal group $O(n)$.

Solution: Let $A$ be a matrix in ${\bf o}(n)$. Then there is a path $\Phi(t)$ in $O(n)$ such that $\Phi(0)=I$ and $\Phi^\prime(0)=A$. That $\Phi(t)$ is orthogonal means $\Phi(t)\Phi(t)^\top=I$. Differentiate both sides, applying the product rule: $\Phi^\prime(t)\Phi(t)^{\top}+\Phi(t)\Phi^{\prime}(t)^\top=O$. Next, evaluate at $t=0$, taking $\Phi(0)=I$ and $\Phi^\prime(0)=A$ into account: $A+A^\top=O$, that is, $A$ is skew symmetric. Conversely, given a skew symmetric matrix $A$ ($A^\top=-A$), consider the path $\Phi(t)=e^{At}$. Since $\Phi(t)$ is the solution to the initial value problem (8.2) above, we have $\Phi(0)=I$ and $\Phi^{\prime}(0)=A\Phi(0)=A$. Thus, in order to show that $A$ belongs to the Lie algebra of $O(n)$, it is enough to check that $\Phi(t)$ belongs to $O(n)$. Indeed,

$$\begin{eqnarray*} \Phi(t)^\top &=&\left({\sum}_{n=0}^\infty\ A^n\frac{t^n}{n!}... ...)^n\frac{t^n}{ n!} \\ &=&e^{-At}=(e^{At})^{-1}=\Phi(t)^{-1}, \end{eqnarray*}$$

showing that $\Phi(t)$ is an orthogonal matrix, or $\Phi(t)\in O(n)$. We conclude: the Lie algebra ${\bf o}(n)$ of the orthogonal group $O(n)$ consists of all skew symmetric $n\times n$ real matrices.

Exercises

1. In each of the following parts, find the matrix differential $d\Phi$ of the given matrix functions $\Phi$ and then evaluate $d\Phi$ at the origin:
   
   $${\rm (a)}\ \ \Phi(t)=\left[\begin{array}{cc} \cosh t & \sinh... ...[\begin{array}{cc} u^2 & uv \\ vu & v^2 \end{array}\right],$$
   
   
   
   
   $${\rm (c)} \ \ \Phi(u,v)=\left[\begin{array}{ccc} \sin u\sin ... ...\\ \sin u\cos v & \cos u\cos v & -\sin v \end{array}\right].$$
   
   
   (Recall the hyperbolic functions $\cosh t=(e^t+e^{-t})/2$ and $\sinh t=(e^t-e^{-t})/2$ and their derivatives: $\cosh^\prime=\sinh$ and $\sinh^\prime=\cosh$.)

   
   

2. Is it true that for every matrix function $\Phi$, we have $d(\Phi^2)=2\Phi d\Phi$? Why?

   
   

3. By a 1-parameter group in a matrix group $G$ we mean a parametric family $\Phi(t)$ ( $-\infty<t<\infty$) in $G$ such that $\Phi(0)=I$ and $\Phi(s+t)=\Phi(s)\Phi(t)$. Verify that $\Phi(-t)=\Phi(t)^{-1}$ and $\Phi^{\prime}(t)=A\Phi(t)$ for all $t$ where $A=\Phi^\prime(0)$.

   
   

4. Verify that, for every invertible matrix function $\Phi$, we have
   
   $$d(\Phi^{-2})=-\Phi^{-2}.\ d\Phi.\ \Phi^{-1}-\Phi^{-1}.\ d\Phi.\ \Phi^{-2}.$$
   
   

   
   

5. Verify by induction that, for every matrix function $\Phi$ and every positive integer $n$, we have $d(\Phi^{n+1})={\sum}_{k=0}^n \Phi^{n-k}.\ d\Phi. \ \Phi^k$. (We adopt the usual convention $\Phi^0=I$.)

   
   

6. Find the matrix exponential function $e^{At}$ for each of the following matrices $A$:
   
   $${\rm (a)} \ \ \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{... ... \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right],$$
   
   
   
   
   $${\rm (e)} \ \ \left[\begin{array}{rr} 1 & 1 \\ 0 & -1 \end... ... & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right].$$
   
   

   
   

7. Given an $n\times n$ matrix $A$, consider the vector fields $\bf F$ in ${\bf R}^n$ defined by ${\bf F}\, \big\vert _{\bf x}=A{\bf x}$; (according to our convention described after Example 8.2 in the present section, a point ${\bf x}$ in ${\bf R}^n$ is considered as a column vector of size $n\times 1$ and thus $A{\bf x}$, interpreted as the product of an $n\times n$ matrix and an $n\times 1$ matrix, is also a point in ${\bf R}^n$). Let $\{\Phi_t\}$ be the flow generated by ${\bf F}$, ${\bf x}_0$ be a point in ${\bf R}^n$ and ${\bf x}={\bf x}(t)$ be a parametric curve in ${\bf R}^n$. Check that the following conditions are equivalent:
   (a) ${\bf x}(t)=\Phi_t({\bf x}_0)$ for all $t$.
   (b) ${\bf x}={\bf x}(t)$ is the solution to the initial value problem $${d{\bf x}\over dt}=A{\bf x}$$, ${\bf x}(0)={\bf x}_0$.
   (c) ${\bf x}(t)=e^{At}{\bf x_0}$.

   
   

8. Solve the following systems of equations and use your answers to find $e^A$ for those $A$ given by parts from (a) to (f) in Exercise 6:
   (a) $dx/dt=y$, $dy/dt=0$ (b) $dx/dt=x+y$, $dy/dt=0$
   (c) $dx/dt=x+y$, $dy/dt=x+y$ (d) $dx/dt=y$, $dy/dt=x$
   (e) $dx/dt=x+y$, $dy/dt=-y$ (f) $dx/dt=y+z$, $dy/dt=z$, $dz=0$.

   
   

9. Use your answer in part (g) of Excercise 6 above to solve the following system of linear differential equations $dx/dt=y$, $dy/dt=-x+z$, $dz/dt=-y$.

   
   

10. Verify that, for $n\times n$ matrices $A$, $B$ and $C$, we have the following Jacobi's identity: $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=O$.

    
    

11. Simplify the expression ${\sum}_{k=0}^n\ A^{n-k}[A,B]B^k$.

    
    

12. Consider the following $3\times 3$ skew matrices
    
    $$J_1=\left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & -1 \\ ... ...0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right].$$
    
    
    Find $e^{tJ_k}$ ($k=1,2,3$) and $[J_1,J_2]$, $[J_2,J_3]$ and $[J_3,J_1]$.

    
    

13. Recall from linear algebra that, given a basis ${\cal E}=\{{\bf e}_1, {\bf e}_2, \ldots, {\bf e}_m\}$ of a vector space $V$, each vector $\bf v$ in $V$ associates with $[{\bf v}]\equiv[{\bf v}]_{\cal E}=(v_1 \ v_2 \ \cdots \ v_m)$ such that ${\bf v}={\sum}_{k=1}^m\, v_k{\bf e}_k$. Also, each linear operator $T$ on $V$ is associated with a matrix $[T]\equiv [T]_{\cal E}$ such that $[T{\bf v}]=[T][{\bf v}]$ for all ${\bf v}$ in $V$. Now consider the adjoint representattion of a matrix Lie group $G$: for each $S$ in $G$, ${\rm Ad}(S)$ is a linear operator on the Lie algebra ${\rm L}G$. Thus we may consider the matrix $[{\rm Ad}(S)]$ if a basis ${\bf E}=\{E_1,\ldots, E_m\}$ in ${\rm L}G$ is chosen. In each of the following cases, compute $[{\rm Ad}(S)]$ for a general element $S$ in $G$:

    
    (a) $G={\rm SO}(2)$ and ${\bf E}=\{E\}$, where $E=\left[\begin{array}{rr} 0 &-1 \\ 1 & 0 \end{array}\right]$. Take $S=\left[\begin{array}{rr} \cos t & -\sin t \\ \sin t & \cos t \end{array}\right]$.

    
    (b) $G={\rm SO}(3)$ and ${\cal E}=\{J_1,\ J_2,\ J_3\}$, (for $J_k$, see Exercise 10 above).

    (c) $G={\rm SL}(2, {\bf R})$, which consists all $2\times 2$ (real) matrices $A$ with $\det(A)=1$, and ${\cal E}=\{E_1,E_2,E_3\}$, where
    
    $$E_1=\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\rig... ...=\left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right].$$