The Telegraph Equation

[durandal]

Homework Statement

Find the general solution of the system of telegraphers equations using the eigenvalues and -vectors.

Relevant Equations

See image in description

I have tried row reduction to solve for the first eigenvector but I don't feel like I get any closer to the solution:

 

[DaveE]

Looks great so far, you just need to keep going.

An important thing to remember about eigenvectors is that there isn't a single vector solution. Eigenvectors are the vectors that satisfy $Av=\lambda v$ so any vector in the same direction as $v$ is also an eigenvector. This means that there is an unspecified degree of freedom that corresponds to the length of the eigenvectors. So, you get to arbitrarily choose one value. You're not solving for a vector, you're solving for a set of vectors of arbitrary magnitude. In practice we only pick and show one, which can confuse people.

ex. if your system has $V_1=[1, -1]^T$ as an eigenvector, then $V_1=[-13, 13]^T$ is also an eigenvector. We would normally say it's the same eigenvector, but that is sloppy shorthand for it's a member of that set of vectors (the eigenspace).

https://www.khanacademy.org/math/li...-finding-eigenvectors-and-eigenspaces-example

PS: Thank you for clearly stating the problem and your attempt. It makes it easy for us to understand and comment. If it's not easy to understand, people might not respond. This is a great example of how to ask a technical question.

 

[durandal]

Looks great so far, you just need to keep going.

An important thing to remember about eigenvectors is that there isn't a single vector solution. Eigenvectors are the vectors that satisfy $Av=\lambda v$ so any vector in the same direction as $v$ is also an eigenvector. This means that there is an unspecified degree of freedom that corresponds to the length of the eigenvectors. So, you get to arbitrarily choose one value. You're not solving for a vector, you're solving for a set of vectors of arbitrary magnitude. In practice we only pick and show one, which can confuse people.

ex. if your system has $V_1=[1, -1]^T$ as an eigenvector, then $V_1=[-13, 13]^T$ is also an eigenvector. We would normally say it's the same eigenvector, but that is sloppy shorthand for it's a member of that set of vectors (the eigenspace).

https://www.khanacademy.org/math/li...-finding-eigenvectors-and-eigenspaces-example

PS: Thank you for clearly stating the problem and your attempt. It makes it easy for us to understand and comment. If it's not easy to understand, people might not respond. This is a great example of how to ask a technical question.

Thank you very much. I found the eigenvectors to be v1 = [1 , -Zc^-1] and v2 = [1 , Zc^-1]. My professor verified it, but he also said that the eigenvectors could be found just by "looking" at the matrix itself. I have asked him about this, but he hasn't answered me. Do you have any idea which "technique" he was referring to?

 

[DaveE]

Thank you very much. I found the eigenvectors to be v1 = [1 , -Zc^-1] and v2 = [1 , Zc^-1]. My professor verified it, but he also said that the eigenvectors could be found just by "looking" at the matrix itself. I have asked him about this, but he hasn't answered me. Do you have any idea which "technique" he was referring to?

About 30 years ago I could have. It's really just proficiency with linear algebra and recognizing "simple" problems. I think the answer lies in understanding the "geometry" associated with linear transforms.

There are some great Linear Algebra lectures online from MIT. Particularly this one which sort of relates to your question.
https://ocw.mit.edu/courses/18-06-l...rces/lecture-21-eigenvalues-and-eigenvectors/

 

[DaveE]

$\begin{bmatrix} \gamma & -Z \\ -Y & \gamma \end{bmatrix} = \gamma \begin{bmatrix} 1 & \frac{-Z}{\gamma} \\ \frac{-Y}{\gamma} & 1 \end{bmatrix} = \gamma \begin{bmatrix} 1 & \frac{-Z}{\sqrt{YZ}} \\ \frac{-Y}{\sqrt{YZ}} & 1 \end{bmatrix} = \gamma \begin{bmatrix} 1 & -\sqrt{\frac{Z}{Y}} \\ -\sqrt{\frac{Y}{Z}} & 1 \end{bmatrix} = \gamma \begin{bmatrix} 1 & -Z_c \\ -\frac{1}{Z_c} & 1 \end{bmatrix}$

The e-vectors are obvious for the matrix on the left. But it's not obvious to me with the version on the right.