Math 309B/C Winter 2012, Homework 3
Due February 1
Recall that if $z=a+ib$, then the real part of $z$ is $a$
and the imaginary part of $z$ is $b$. The real and
imaginary parts are both real.
-
- Show that $\overline{e^{a+ib}}=e^{a-ib}$ (Hint:
recall that $e^{a+ib}=e^{a}(\cos b+i\sin b)$. (We will need
this identity a lot.)
- Show that $z\overline{z}=|z|^2$. Recall that if $z=a+ib$,
then $|z|=\sqrt{a^{2}+b^{2}}$.
- Show that $\overline{\overline{z}}=z$.
- Show that if $z=a+ib$ with $b\neq 0$, then $z$ and
$\overline{z}$ are linearly independent with respect to the
real numbers, i.e. there are no nonzero real constants $c_1$ and
$c_2$ such that $c_1z+c_2\overline{z}=0$. Hint:
Write $z$ as $z=a+ib$ where $a$ and $b$ are real numbers.
\[c_1z+c_2\overline{z}
=c_1(a+ib)+c_2(a-ib)=(c_1+c_2)a+i(c_1-c_2)b=0.\] If you look
at the real and imaginary parts, those have to both equal
zero, so we have the system of equations \[ac_1+ac_2=0\]
\[bc_1-bc_2=0,\] i.e. we're solving the matrix equation
\[\left(\begin{array}{cc} a & a \\ b & -b
\end{array}\right) \left(\begin{array}{c} c_1 \\ c_2
\end{array}\right)=0.\] So now we just have to show that any
solution $\left(\begin{array}{c} c_1 \\ c_2
\end{array}\right)=0$ to this equation must be zero. Hint:
look at the determinant.
- Let $A$ be an $n\times n$ matrix with real
entries.
- Show that if $v$ is a complex eigenvector with eigenvalue
$\lambda$, then $\overline{v}$ (i.e. the conjugate of the
vector) is an eigenvector with eigenvalue
$\overline{\lambda}$. (What this exercise says is that
whenever you have an eigenvalue, it's complex conjugate is
also an eigenvalue. This can be helpful when trying to solve
for eigenvalues, since if you find one complex one, you've
automatically found a second one. In other words, complex
eigenvalues and eigenvectors come in conjugate pairs.)
Answer: Since $v$ is an eigenvector, $Av=\lambda v$.
Conjugating both sides of the equation, we get
$\overline{Av}=\overline{\lambda v}$. Now notice that
$\overline{A}=A$ since $A$ is real, hence
$\overline{Av}=A\bar{v}=\overline{\lambda
v}=\bar{\lambda}\bar{v}$, but this is what it means for
$\bar{v}$ to be an eigenvector with eigenvalue
$\bar{\lambda}$.
- Show that if $v=\left(\begin{array}{c} a+ib \\
c+id\end{array}\right)$ is a complex eigenvector for $A$ (a
2x2 matrix) with complex eigenvalue $\lambda$, then
$ad-bc=0$. Hint: If $v$ is an eigenvector with
e-value $\lambda$, then so is $(a-ib)\cdot v$. Compute what
this vector is and note that if $ad-bc=0$, then $(a-ib)v$ is
a real eigenvector with a complex eigenvalue, but this can't
happen since $A$ is real.
Answer: Note that \[(a-bi)\cdot v=
\left(\begin{array}{c} a^{2}+ b^{2} \\
ac+bd+i(ad-bc)\end{array}\right).\] If $ac-bd=0$, then
$(a-bi)\cdot v$ is real, and so $A((a-ib)v)$ is real (since
$A$ is real), but $A((a-bi)\cdot v)=\lambda(a-bi)\cdot v$,
which is complex (since $\lambda$ is complex), hence
$ad-bc\neq 0$.
- Suppose $x=a+ib$ is now a vector, with $a$ and $b$ real
vectors. Then $x$ and $\overline{x}$ are linearly independent
with respect to the reals. (This follows from problem 1d,
although you don't need to prove it, but you will need to know
this fact.)
- Show that if $x(t)$ solves $x'=Ax$, where $A$ is a real
matrix, then
- $\overline{x}(t)$ also solves this equation.
Answer: Note that if $x'=Ax$, then
$\bar{x'}=\overline{Ax}=A\bar{x}$ since $A$ is real.
- Suppose $A$ is a $2\times 2$ matrix, and
$x(t)=ve^{\lambda t}$ is a solution with $\lambda=a+ib$
complex (i.e. $b\neq 0$). Show that $x(t)$ and
$\overline{x}(t)$ are linearly independent for all $t$.
Hint: You just need to show that the Wronskian is
nonzero somewhere. Let $v=\begin{array}{c}
\alpha+i\beta \\ \gamma + i\rho\end{array}$ and compute
$\det(v|\overline{v})$. By problem 2b, this will be nonzero
at $t=0$, why?
Answer: We just need to show that the Wonskian is
nonzero at $t=0$. Writing $x(0)=v=\left(\begin{array}{c}
a+ib \\ c+id\end{array}\right)$,
\[W(0)=\det(x(0)|\bar{x}(0))=\det \left(\begin{array}{cc}
a+ib & c+id \\ a-ib & c-id \end{array}\right) =
2i(bc-ad),\] and recall that $v$ is an eigenvector, so by
problem 2b, $bc-ad\neq 0$, so $W(0)\neq 0$.
Recall that, when we were solving $x'=Ax$, we functions of the form
$x(t)=ve^{\lambda t}$, where $v$ is an eigenvector and $\lambda$ is
its eigenvalue, are solutions. The same still holds if we get
complex eigenvalues, so solving these systems is almost the same as
before, but now (by the previous problem) we know that they come in
conjugate pairs, i.e., $\overline{v}e^{\overline{\lambda}t}$ is also
a solution..
-
- Solve $x'=\left(\begin{array}{cc} 1 & 1 \\ -1 & 1
\end{array}\right)x$ for its general solutions. You should
get complex eigenvalues and complex solutions
\[x^{(1)}=\left(\begin{array}{c} 1 \\ i \end{array}\right)
e^{(1+i) t} \;\;\; \mbox{ and } \;\;\;
x^{(2)}=\left(\begin{array}{c} 1 \\ -i \end{array}\right)
e^{(1-i) t},\] or perhaps multiples of these.
You solve these equations exactly the same way as in the
previous section, only these complex roots show up. Hint:
By the previous problem, if you ever solve an equation of the
form $x'=Ax$ and get a complex solution $x$, then
$\overline{x}$ is another fundamendal solution. Hence, if $A$
is $2\times 2$, if you find one fundamental solution, then
it's a synch to find the other one...just conjugate the first
one! Notice, in particular, that the solutions in the previous
problem are complex conjugates of each other.
We could consider ourselves done in solving this problem, but
in real life applications we typically are dealing with real
functions and real initial conditions, and should expect that
our solutions are also real. Hence, it is important to know
how to express our general solutions interms of real
fundamental solutions.
- With the solution $x^{(1)}$ from the previous problem,
compute the real and imaginary parts of this vector
function. You will then get two real-valued vector functions
$y^{(1)}=\left(\begin{array}{c} \cos t \\ -\sin
t\end{array}\right) e^{t}$ and
$y^{(2)}=\left(\begin{array}{c} \sin t \\ \cos
t\end{array}\right) e^{t}$.
- Show that $y^{(1)}$ and $y^{(2)}$ are a fundamendal set
of solutions. Hint: Compute the Wronskian of the
two vectors for some value $t$, say $t=0$.
- So when dealing with $x'=Ax$ where $A$ is a matrix with two
complex eigenvalues $a\pm ib $, to solve for the real valued
general solutions, do the following:
- Solve for the eigenvalues $a\pm ib$ and corresponding
eigenvectors $v_{\pm}$. The fundamental solutions will look
(just as before) \[x^{(1)}= v_1 e^{(a+ib)t} \;\;\; \mbox{
and } \;\;\; x^{(2)}= v_2 e^{(a-ib)t}.\]
- Solve for the real and imaginary parts of one these
fundamental solutions, say $x^{(1)}$, by writing $v_1=u+iw$
(where $u$ and $w$ are real vectors). If you substitute this
in the above expression, write $e^{(a+ib)t}=e^{at}(\cos
bt+i\sin bt)$ and expand, you should get
\[x^{(1)}=e^{at}(u\cos bt-w\sin t)+ie^{at}(u\sin bt + w \cos
bt).\] Then your two real fundamental solutions are the real
and imaginary parts of this expression
\[y^{(1)}=e^{at}(u\cos bt-w\sin t), \;\;\;
y^{(2)}=e^{at}(u\sin bt + w \cos bt).\]
- Note that if the real value is positive (i.e. $a>0$),
then the solution will grow to infinity, and if the real
part is negative, then it will decay to zero. In either
case, we call the origin a spiral point.
Try the following problems to get some practice. Solve for a
real set of fundamental solutions and write the general form of
a real solution (and for fun (i.e. not necessary for the quiz),
try plotting some of your
solutions):
- $x'=\left(\begin{array}{cc} 1 & -3 \\ 3 & 1
\end{array}\right)$. To see why we call these solutions spirals,
here is a plot of a solution to this system:
Answer: This has eigenvalues $\lambda=1+\pm 3$. To find
the eigenvectors, note that \[(A-(1+3i)I)x =
\left(\begin{array}{cc} -3i & -3 \\ 3 & -3i
\end{array}\right)\begin{array}{c} x_{1} \\ x_{2}
\end{array}\right).\] Note that if $x$ is an eigenvector, then
so is any constant multiple. Hence, we can actually pick $x_1$
to be whatever we want, so let's pick it to be some thing so
that, when we compute the resulting vector above, there is no
$i$, so let $x_1=i$. Then $x_2=1$ by setting the above equation
equal to zero and solving for $x_2$. That is the first
eigenvector, and by one of our earlier problems, the eigenvector
for $1-3i$ is just the conjugate of this one. Hence, our first
eigenvector is $\left(\begin{array}{c} \pm i \\ 1
\end{array}\right)$. So one of your fundamental solutions is
\[x^{(1)}(t)= \left(\begin{array}{c} i \\ 1 \end{array}\right)
e^{(1+3i)t}.\] Let's compute the real and imaginary parts of
this: \[x^{(1)}(t) = \left[ \left(\begin{array}{c} 0 \\ 1
\end{array}\right) + i \left(\begin{array}{c} 1 \\ 0
\end{array}\right) \right] e^{t}(\cos 3t+i\sin 3t) \] \[=
\underbrace{e^{t}\left[ \left( \begin{array}{c} 0 \\ 1
\end{array}\right)\cos t - \left(\begin{array}{c} 1 \\ 0
\end{array}\right)\sin t\right]}_{y^{(1)}}
+\underbrace{e^{t}\left[\left( \begin{array}{c} 0 \\ 1
\end{array}\right)\sin t + \left(\begin{array}{c} 1 \\ 0
\end{array}\right)\cos t\right]}_{y^{(2)}},\] and $y^{(1)}$ and
$y^{(2)}$ are your real fundamdental solutions, and the general
form is \[ c_{1} e^{t}\left[ \left(\begin{array}{c} 0 \\ 1
\end{array}\right)\cos t - \left(\begin{array}{c} 1 \\ 0
\end{array}\right)\sin t\right] + c_{2} e^{t}\left[
\left(\begin{array}{c} 0 \\ 1 \end{array}\right)\sin t +
\left(\begin{array}{c} 1 \\ 0 \end{array}\right)\cos t\right],\]
- $x'=\left(\begin{array}{cc} 3 & -2 \\ 4 & -1
\end{array}\right)x$. \[x= c_{1} e^{t} \left(\begin{array}{c}
\cos 2t \\ \cos 2t+\sin 2t \end{array}\right)
+c_{2}e^{t}\left(\begin{array}{c} \sin 2t \\ -\cos 2t +\sin
2t\end{array}\right).\]
- $x'=\left(\begin{array}{cc} -1 & -4 \\ 1 & -1
\end{array}\right)x.
- $x'=\left(\begin{array}{cc} 2 & -5 \\ 1 & -2
\end{array}\right)x$
Answer The eigenvalues are $\lambda =\pm 1$. To solve
for the eigenvectors, let $\lambda =i$, and solve
\[0=(A-iI)x=\left(\begin{array}{cc} 2-i & -5 \\ 1 & -2-i
\end{array}\right)\left(\begin{array}{c} x_1 \\
x_2\end{array}\right).\] Again, let us pick $x_1$ so that when
it gets multiplied by the top left entry in the matrix (i.e.
2-i), it becomes real, so let $x_1=2+i$ (i.e. the conjugate of
$2-i$). Then the above equals \[\left(\begin{array}{cc} 5-5x_2
\\ 2+i +(-2-i)x_2\end{array}\right),\] and since this all equals
zero, we should pick $x_2=1$, and so our eigenvector for
$\lambda =i$ is $\left(\begin{array}{c} 2+i \\ 1
\end{array}\right)$, and the eigenvector for $-i$ is just the
conjugate of this. Our first fundamental solution is
\[x^{(1)}(t)= \left(\begin{array}{c} 2+i \\ 1 \end{array}\right)
e^{it}.\] Let's compute the real and imaginary parts of this:
\[x^{(1)}(t) = \left[ \left(\begin{array}{c} 2 \\ 1
\end{array}\right) + i \left(\begin{array}{c} 1 \\ 0
\end{array}\right) \right] (\cos t+i\sin t) \] \[=
\underbrace{\left[ \left( \begin{array}{c} 2 \\ 1
\end{array}\right)\cos t - \left(\begin{array}{c} 1 \\ 0
\end{array}\right)\sin t\right]}_{y^{(1)}}
+\underbrace{\left[\left( \begin{array}{c} 2 \\ 1
\end{array}\right)\sin t + \left(\begin{array}{c} 1 \\ 0
\end{array}\right)\cos t\right]}_{y^{(2)}},\] and $y^{(1)}$ and
$y^{(2)}$ are your real fundamdental solutions, and the general
form is \[ c_{1} \left[ \left(\begin{array}{c} 2 \\ 1
\end{array}\right)\cos t - \left(\begin{array}{c} 1 \\ 0
\end{array}\right)\sin t\right] + c_{2} \left[
\left(\begin{array}{c} 2 \\ 1 \end{array}\right)\sin t +
\left(\begin{array}{c} 1 \\ 0 \end{array}\right)\cos t\right].\]
- $x'=\left(\begin{array}{cc} 3 & -2 \\ 4 & -1
\end{array}\right)$
- This problem deals with a system $x'=Ax$ where the
eigenvalues of $A$ have no real part, i.e. they are $\pm ib$ for
some real number $b$. The solution to such an equation will just
spiral around in a circle: it won't diverge to infinity or
converge to the origin, but will orbit $0$. In such a case, we
call the origin a center. Solutions are stable in the
sense that they don't diverge to infinity, but they are
considered asymptotically unstable because they never converge
to any vector.
- Solve $x'=\left(\begin{array}{cc} 0 & 1 \\ -1 & 0
\end{array}\right)x$ for its fundamental solutions.
- Convert the fundamental solutions into real ones by
computing the real and imaginary parts of one of the
fundamental solutions. In the end, you should get
fundamental solutions of the form
\[y^{(1)}(t)=\left(\begin{array}{c} \cos t \\ -\sin
t\end{array}\right), \;\;\;
y^{(2)}(t)=\left(\begin{array}{c} \sin t \\ \cos
t\end{array}\right).\] If you didn't get these exactly, your
solution is still correct if your fundamental solutions were
multiples of these ones. Note that there is no exponential
term, so any linear combination of these guys (i.e. a
general solution) won't diverge to infinity or converge to
the origin.
- In this problem, you'll prove that any general solution
will just spiral around in a circle. Show that if \[x(t)=
c_{1} \left(\begin{array}{c} \cos t \\ -\sin
t\end{array}\right)+ c_2 \left(\begin{array}{c} \sin t \\
\cos t\end{array}\right),\] then
\[x(t)=c\left(\begin{array}{c} \\cos(t+\theta) \\
-\sin(t+\theta)\end{array}\right),\] where
\[c=\sqrt{c_1^2+c_2^2}, \;\;\; \theta = \arccos
\frac{c_1}{c}.\] Hint: Start from the right side of
the equation and espand using angle formulas for sin and
cos, and remember that if $\theta=\arccos t$, then $\sin
t=\sqrt{1-t^2}$.
Answer: Note that $\arcsin \theta = \sqrt{
1-\frac{c_{1}^{2}}{c^{2}}}=\sqrt{1-\frac{c_{1}^{2}}{c_{1}^{2}+c_{2}^{2}}}=\sqrt{\frac{c_{2}^{2}}{c^{2}}}=\frac{c_{2}}{c}$,
soStart by expanding what we think the solution is:
\begin{align*} x(t) & = c\left(\begin{array}{c}
\cos(t+\theta) \\ -\sin(t+\theta)\end{array}\right)
=c\left(\begin{array}{c} \cos t \cos \theta -\sin t\sin
\theta \\ -\sin t\cos \theta -\sin\theta\cos t
\end{array}\right) \\ & =c\left(\begin{array}{c} \cos t
\frac{c_{1}}{c} -\sin t\frac{c_{2}}{c} \\ -\sin
t\frac{c_{1}}{c}-\frac{c_{2}}{c}\cos t \end{array}\right)\\
& =\left(\begin{array}{c} c_1\cos t -c_2\sin t \\
-c_1\sin t-c_2\cos t \end{array}\right) \\ & =c_{1}
\left(\begin{array}{c} \cos t \\ -\sin t\end{array}\right)+
c_2 \left(\begin{array}{c} \sin t \\ \cos
t\end{array}\right).\end{align*}
- Show that $x(t)$ travels in a circle around the origin.
What's its radius? Hint: If you're moving around in
a circle around the origin, then your distance to zero is
constant (imagine someone swinging $x(t)$ around the origin
with a rope, then it orbits the point in a circle of radius
equal to the length of the rope). Hence, you only need to
show that the norm of $x(t)$ is constant (and it's norm will
be the radius of the circle).
Suppose now that you have a 3x3 real matrix $A$ that has a complex
eigenvalue $\lambda$ with a complex eigenvalue $\lambda$ with
eigenvector $v$. Again, we'd like to express the general solution
for this equation in terms of real fundamental solutions.
By problem 2, $\overline{\lambda}$ is also an eigenvalue with
eigenvector $\overline{v}$. Then the third eigenvalue $\mu$ must be
real (otherwise, $\overline{\mu}$ would also be an eigenvalue, but
then we'd have 4 eigenvalues, which is impossible for a 3x3 matrix)
and has a real eigenvector $u$. Then the fundamental solutions are
$ue^{\mu t}, ve^{\lambda t},$ and
$\overline{v}e^{\overline{\lambda}t}$. The general solution will (as
always) look like \[x(t)=c_1 ue^{\mu t}+c_2 ve^{\lambda t}+c_3
\overline{v}e^{\overline{\lambda}t}.\] Again, we'd like to have a
general solution in terms of real fundamental solutions.
While the first term in our general solution (the one with the
$\mu$) is ok, since everything is real, we just have to replace the
other two vectors with real functions. You do the same trick as
before: take the real and imaginary parts of one of the complex
fundamental solutions (say $ve^{\lambda t}$), decompose it into its
real and imaginary parts to get two new fundamental solutions
$y^{(1)}$ and $y^{(2)}$. Hence, a general solution with real
fundamental solutions will have the form \[c_1 ue^{\mu t}+c_2
y^{(1)}+c_3 y^{(2)}.\] Express the general solutions to the
following systems in terms of real fundamental solutions:
- $x'=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1
& −2 \\ 3 & 2 & 1\end{array}\right)x$.
- $x'=\left(\begin{array}{ccc} -3 & 0 & 2 \\ 1 & -1
& 0 \\ -2 & -1 & 0\end{array}\right)x$.
- The following problems deal with repeated eigenvalues. Solve
for the general solutions of each system, and solve the initial
value problem if there is an initial condition.
- $x'=\left(\begin{array}{cc} 3 & -4 \\ 1 & -1
\end{array}\right)x$
- $x'=\left(\begin{array}{cc} 4 & -2 \\ 8 & -4
\end{array}\right)x$
- $x'=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 2 &
1 & -1 \\ 0 & -1 & 1 \end{array}\right)x$
- $x'=\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 &
0 & 1 \\ 1 & 1 & 0 \end{array}\right)x$
- $x'=\left(\begin{array}{cc} 1 & -4 \\ 4 & -7
\end{array}\right)x$, $x(0)=\left(\begin{array}{c} 3 \\ 2
\end{array}\right)$
Answer: This has eigenvalue $\lambda =-3$ with
multiplicity 2 and eigenvector $v=\left(\begin{array}{c} 1
\\ 1 \end{array}\right)$. One fundamental solution is
\[x^{(1)}=ve^{\lambda t}=\left(\begin{array}{c} 1 \\ 1
\end{array}\right)e^{-3t}.\] To solve for the other one,
recall that \[x^{(2)}=ue^{-3t}+vte^{-3t}\] where $v$ is out
eigenvector and $u$ satisfies \[(A-\lambda I)u=v,\] i.e.
\[(A-(-3)I)u=\left(\begin{array}{c} 1 \\ 1
\end{array}\right).\] Solving this equation gives
$u=\left(\begin{array}{c} 1 \\ 1
\end{array}\right)k+\left(\begin{array}{c} 1/4 \\ 0
\end{array}\right)$ for some arbitrary constant $k$. Since
we just want one solution, just pick $k=0$, so
$u=\left(\begin{array}{c} 1/4 \\ 0 \end{array}\right)$, and
our general solution is now \[x(t)= c_{1}
x^{(1)}+c_{2}x^{(2)}= c_{1} \left(\begin{array}{c} 1 \\ 1
\end{array}\right)e^{-3t}+ c_{2} \left(
\left(\begin{array}{c} 1/4 \\ 0 \end{array}\right)+
\left(\begin{array}{c} 1 \\ 1
\end{array}\right)te^{-3t}\right)\] To solve the initial
value problem, we set \[\left(\begin{array}{c} 3 \\ 2
\end{array}\right)= x(0)=c_{1} \left(\begin{array}{c} 1 \\ 1
\end{array}\right)+ c_{2} \left(\begin{array}{c} 1/4 \\ 0
\end{array}\right)= \left(\begin{array}{cc} 1 & 1/4 \\ 1
& 0 \end{array}\right)\left(\begin{array}{c} c_{1} \\
c_{2}\end{array}\right),\] and now we solve this equation
(which is just linear algebra at this point).
- $x'=\left(\begin{array}{cc} 2 & 3/2 \\ -3/2 & -1
\end{array}\right)x$, $x(0)=\left(\begin{array}{c} 3 \\ -2
\end{array}\right)$
- $x'=\left(\begin{array}{cc} 3 & 9 \\ -1 & -3
\end{array}\right)x$, $x(0)=\left(\begin{array}{c} 2 \\ 4
\end{array}\right)$
- Consider the equation $x'=\left(\begin{array}{ccc} 1 & 1
& 1 \\ 2 & 1 & -1 \\ -3 & 2 & 4
\end{array}\right)x$. Compute all eigenvalues and eigenvectors
(there should be only one of each, call them $\lambda$ and $v$
respectively). Here's how to find some fundamental solutions
$x^{(1)},x^{(2)},x^{(3)}$:
- The first one is just $x^{(1)}=ve^{\lambda t}$ as usual.
- Let $x^{(2)}=v_1 e^{\lambda t}+v_2 t e^{\lambda t}$,
solve for $v_1$ and $v_2$.
- Let $x^{(3)}=u_1 e^{\lambda t} + u_2 t e^{\lambda t}+ u_3
\frac{t^{2}}{2} e^{\lambda t}$, solve for $u_1, u_2,$ and
$u_3$.
For the following systems, find the fundamental matrix $\Phi$
(i.e. the matrix s.t. $\Phi'=A\Phi$) such that $\Phi(0)=I$, and
then solve the initial value problems $x(0)=e_{1}$ and
$x(0)=e_{2}$ (where $e_{1}$ and $e_{2}$ are standard basis
vectors):
- $x'=\left(\begin{array}{cc} 3 & -2 \\ 2 & -2
\end{array}\right)x$
- $x'=\left(\begin{array}{cc} 2 & -1 \\ 3 & -2
\end{array}\right)x$
- $x'=\left(\begin{array}{cc} 2 & -5 \\ 1 & -2
\end{array}\right)x$ (Note that you'll get complex
eigenvalues, first find some real fundamental
solutions to make your fundamental matrix.)
- Show that if $\Phi(t)$ satisfies $\Phi'=A\Phi$ and
$\Phi(0)=I$, then $\Phi(t)\Phi(s)=\Phi(t+s)$. (Hint: Fix $s$.
Show that, for any vector $v$, $x(t)=\Phi(t)\Phi(s)v$ and
$x(t)=\Phi(t+s)v$, satisfy the same initial value problem
$x'=Ax$ and $x(0)=\Phi(s)v$. Now apply uniqueness, and recall
that two matrices $M$ and $N$ are equal if and only if $Mv=Nv$
for all vectors $v$.)
Answer: Fix $s$ (that is, suppose it's some number that's not
varying, this means that when we differentiate, we don't
differentiate in $s$, only $t$). Note that
\[x(t)=\Phi(t)\Phi(s), \;\;\; y(t)= \Phi(t+s)\] both solve the
initial value problem $X'=AX$ with initial data $X(0)=\Phi(s)$.
Hence, by the uniqueness theorem, they are the same function.