The question asks to find the solution of the system x'(t)=Ax(t)

given my A is
$\begin{bmatrix}
-7 & 0 & 6 \\
0 & 5 & 0 \\
6 & 0 & 2
\end{bmatrix}
$

After finding my det( $\lambda I -A)$=> $\lambda ^3 -75\lambda +250 =0$

I found three real roots:
$\lambda 1=-10$
$\lambda 2=5$
$\lambda 3=5$

Now starting with the first root equal to -10 and plugging into the det matrix
$\begin{bmatrix}
-3 & 0 & -6 \\
0 & -15 & 0 \\
-6 & 0 & -12
\end{bmatrix}
$

*
$\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
$

=
$\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}
$

I found my U1 to be <2, 0, -1>

Now I'm confused what to do with the last two lambdas, I get the feeling that there should be more than just two Cn's
$x(t)=C1 e^-10t\begin{bmatrix}
2\\
0\\
1
\end{bmatrix}+C2...+C3...
$

My question is can someone show me how to solve for u2 and u3 (if there is a u3), the way my textbook explains solving for eigenvectors is not helping very much considering the questions are getting harder and the method I've been using doesn't seem to be working.

2. Originally Posted by 360modina
The question asks to find the solution of the system x'(t)=Ax(t)

given my A is
$\begin{bmatrix}
-7 & 0 & 6 \\
0 & 5 & 0 \\
6 & 0 & 2
\end{bmatrix}
$

After finding my det( $\lambda I -A)$=> $\lambda ^3 -75\lambda +250 =0$

I found three real roots:
$\lambda 1=-10$
$\lambda 2=5$
$\lambda 3=5$

Now starting with the first root equal to -10 and plugging into the det matrix
$\begin{bmatrix}
-3 & 0 & -6 \\
0 & -15 & 0 \\
-6 & 0 & -12
\end{bmatrix}
$

*
$\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
$

=
$\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}
$

I found my U1 to be <2, 0, -1>

Now I'm confused what to do with the last two lambdas, I get the feeling that there should be more than just two Cn's
$x(t)=C1 e^-10t\begin{bmatrix}
2\\
0\\
1
\end{bmatrix}+C2...+C3...
$

My question is can someone show me how to solve for u2 and u3 (if there is a u3), the way my textbook explains solving for eigenvectors is not helping very much considering the questions are getting harder and the method I've been using doesn't seem to be working.
When you have a matrix equation of the form $\frac{\,d\bold x}{\,dt}=\bold A\bold x$, where $\bold A$ is the coefficient matrix, we assume the solution to be of the form $\bold x=\bold ve^{\lambda t}$

After substitutions and simplifications, we end up with $\left(\bold A-\lambda\bold I\right)\bold v=\bold 0$, where $\bold v$ is the eigenvector associated with the eigenvalue $\lambda$

To find the eigenvalues, we must find where $\det\left(\bold A-\lambda\bold I\right)=0$

Your coefficient matrix is $\bold A=\begin{bmatrix}-7 & 0 & 6 \\ 0 & 5 & 0 \\ 6 & 0 & 2\end{bmatrix}$

Thus, to find the eigenvalues, they must satisfy the equation $\det\left(\bold A-\lambda\bold I\right)=0$

Thus, we see that

$\left|\begin{array}{ccc}-7-\lambda & 0 & 6 \\ 0 & 5-\lambda & 0 \\ 6 & 0 & 2-\lambda\end{array}\right|=0\implies (-7-\lambda)\left|\begin{array}{cc}5-\lambda & 0 \\0 & 2-\lambda\end{array}\right|+6\left|\begin{array}{cc} 0 & 5-\lambda\\ 6 & 0\end{array}\right|=0$

$\implies (-7-\lambda)(5-\lambda)(2-\lambda)-36(5-\lambda)=0\implies (5-\lambda)\left[(-7-\lambda)(2-\lambda)-36\right]=0$

$\implies (5-\lambda)\left[-50+5\lambda+\lambda^2\right]=0\implies(5-\lambda)(10+\lambda)(-5+\lambda)=0$

$\implies\begin{array}{rcr}\lambda_1&=&5\\\lambda_2 &=&5\\\lambda_3&=&-10\end{array}$

We the case of repeated eigenvalues.

The eigenvector for $\lambda = -10$ would be found by solving the equation $\begin{bmatrix}3 & 0 & 6 \\ 0 & 15 & 0 \\ 6 & -5 & 12\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bm atrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

You end up with the equations $3v_1+6v_3=0$, $15v_2=0$ and $6v_1-5v_2+12v_3=0$

Since $v_2=0$, we see that $-\tfrac{1}{2}v_1=v_3$

Thus, we can let the eigenvector be $\bold v_3=\begin{bmatrix}2\\0\\-1\end{bmatrix}$

Now, let's look at what happens when $\lambda = 5$

We have the equation

$\begin{bmatrix}-12 & 0 & 6 \\ 0 & 0 & 0 \\ 6 & -5 & -3\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bma trix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

We come up with the equations $-12v_1+6v_3=0\implies 2v_1-v_3=0$ and $6v_1-5v_2-3v_3=0$

Letting $v_2=0$, we end up with $6v_1-3v_3=0\implies 2v_1-v_3=0\implies v_3=2v_1$

We can have one of the eigenvectors being $\bold v_1=\begin{bmatrix}2\\0\\4\end{bmatrix}$

However, I'm not able to see another possible eigenvector!!!

Up to this point, our solution is $\bold x(t)=c_1\begin{bmatrix}2\\0\\4\end{bmatrix}e^{5t}+ c_2\begin{bmatrix}2\\0\\-1\end{bmatrix}e^{-10t}+\dots$

Anyone can chip in, if they'd like...I have a feeling there isn't another eigenvector...

Does this make sense?

--Chris

3. Hello,
Originally Posted by Chris L T521

The eigenvector for $\lambda = -10$ would be found by solving the equation $\begin{bmatrix}3 & 0 & 6 \\ 0 & 15 & 0 \\ 6 & {\color{red}-5} & 12\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bm atrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Maybe I'm just with stupid, but why did you put that red -5 ?
Though it doesn't change the result.

$\begin{bmatrix}-12 & 0 & 6 \\ 0 & 0 & 0 \\ 6 & {\color{red}0} & -3\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bma trix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
We have then the equations

$\left\{\begin{array}{ll} -12 v_1+6v_3=0 \\ 6v_1-3v_3=0 \end{array} \right.$

Okay, solve calmly these equations, you get :
$2v_1=v_3$
So you can take $v_1=1$ and $v_3=2$

Now the difference is in $v_2$.
I don't really remember why, but you have to take $v_2=1$ and $v_2=0$, because of matters of linear independence.
If you think about it, it'd be logic. However, I'm sorry I don't know what the explanation is ><

Letting $v_2=0$
That was your mistake. What if $v_2 \neq 0$ ???
I admit that was not really a mistake. Actually, it could be directly deduced that $v_2=0$ from your equations since $6v_1-3v_3=-3(2v_1-v_3)=0$ and hence $-5v_2=0$.