Question

Find all the solutions to: xdot = [0 1; 0 1] * x

Answer 1

I understand that the first step is to find the eigenvalues, however doing so results in eigenvalues with a multiplicity of two. I'm confused as to how to continue.

Answer 2

Is that
\[\dot{x} = \left[\begin{matrix}0 & 1 \\ 0 & 1\end{matrix}\right]x\]
?

Answer 3

Yes, it is, sorry!

Answer 4

oh I'm sorry

Answer 5

it's actually \[\left[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right]\]

Answer 6

Ah okay, I was gonna say, that's not really your typical matrix problem :)

Answer 7

Yeah, sorry fail haha. My brain is a little off today!

Answer 8

I guess the multiplicity is the part that throws me the most.

Answer 9

So we say that
\[x = \left(\begin{matrix}a_1 \\ a_2\end{matrix}\right)e^{\lambda t} \]
so then
\[\dot{x} = \left[\begin{matrix}\lambda & 0 \\ 0 & \lambda\end{matrix}\right]\left(\begin{matrix}a_1 \\ a_2\end{matrix}\right)e^{\lambda t} \]
so subtracting and taking the determinant, we get
\[\det(\left[\begin{matrix}1-\lambda & 2 \\ 0 & 1-\lambda\end{matrix}\right])= (1-\lambda)^2 = 0\]
so we get degenerate eigenvalues,
\[\lambda = 1\]
plugging that in to our matrix, we see that
\[\left[\begin{matrix}0 & 2 \\ 0 & 0\end{matrix}\right]\left(\begin{matrix}a_1 \\ a_2\end{matrix}\right)=0\]
which means that
\[a_2 =0\]
and so we only get one eigenvalue / eigenvector pair from this, i.e.
\[\lambda = 1, \vec{v} = \left(\begin{matrix}1 \\ 0\end{matrix}\right)\]
So our matrix has only one eigenvector, and the matrix is called defective.

Answer 10

I think you probably did the entire problem right, but then just got hung up on the fact that it doesn't have a complete set of eigenvectors.

Answer 11

But doesn't the multiplicity have to be taken into account? I thought repeated roots had that affect.

Answer 12

This is a first order differential equation. If you look at the equations separately, they are
\[\dot{x_1}(t) = a_1x_1 + 2a_2x_2 \]
and
\[\dot{x_2}(t) = a_2x_2\]

Answer 13

Note that you can solve the second one already, and you get that
\[x_2 = \frac{a_2}{\lambda} e^{\lambda x} \]
plugging that into the first one,
\[\dot{x_1} = a_1x_1 + 2e^{\lambda x} \]

Answer 14

To answer your question more directly, yes, you do need to take the multiplicity into account, but that matrix has only one eigenvector, so the solution up to that point is right.

Answer 15

Alright, so the work I did before reached the same conclusion you have so far. From there, I tried to use the following method: \[\left[ A \right]* \left[ P \right] = \left[ K \right]\]

Where A is the matrix [1 2; 0 1]. and [K] is the first calculated matrix [0 1]. I then solved for the P matrix.

Answer 16

Okay, sounds good. That gives us
\[\left[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right]\left(\begin{matrix}\rho_1 \\ \rho_2\end{matrix}\right)=\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]
so again, the solution is
\[\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]
which means that our final answer is
\[\vec{x_2}(t) = \left(\begin{matrix}1 \\ 0\end{matrix}\right)te^t +\left(\begin{matrix}1 \\ 0\end{matrix}\right)e^t\]

Answer 17

Do you think that method seem correct or am I barking up the wrong tree?

Answer 18

No, you are correct. As an overview, if you have degenerate eigenvalues, you should get the following.
If
\[\dot{\vec{x}}(t) = A\vec{x}(t)\]
and
\[\vec{a}\]
is the eigenvector corresponding to the degenerate eigenvalue lambda, then
\[\vec{x_1}(t) = \vec{a}e^{\lambda t} \]
\[\vec{x_2}(t) = \vec{a}te^{\lambda t} + \vec{b}e^{\lambda t}\]
are the two linearly independent solutions where
\[\vec{b}\]
satisfies the equation
\[A\vec{b} = \vec{a} \]

Answer 19

Alright, thank you very much Joe! I appreciate your help and patience in this sir.

Answer 20

No problem, good luck with your class