Question

Theoretical example system of DEs with only 1 eigenvector and eigenvalues mu_g(lambda) 11 years ago

Answer 1

\[x ^{'}(t)=\left[\begin{matrix}1 & -1 \\ 1 & 3\end{matrix}\right]x(t)\]

eigenvalue = 2
eigenvector = (1,-1)

\[\rightarrow x _{1}(t)=\left(\begin{matrix}1 \\ -1\end{matrix}\right)e ^{2t}\]

Problem: only one apparent solution where we expect 2.
Simple calculus method: replace e^2t with te^2t and find solution to

\[x _{2}(t)=\eta te ^{2t}\]
\[x^{'} _{2}(t)=2\eta te ^{2t} + \eta e ^{2t}\]
\[\rightarrow 2\eta te ^{2t} + \eta e ^{2t} - \left[\begin{matrix}1 & -1 \\1 & 3\end{matrix}\right]\eta te ^{2t}=0\]
which could only work if eta = 0

Since the last equation holds both e^2t as well as te^2t we could combine them and say that
\[x _{2}(t)=\eta te ^{2t} + \zeta e ^{2t}\]
after taking the derivative that would form the following equation

\[2\eta te ^{2t}+ (\eta + 2\zeta) e ^{2t}=\left[\begin{matrix}1 & -1 \\ 1 & 3\end{matrix}\right](\eta te ^{2t} + \zeta e ^{2t})\]

I follow this far. But then my course says "If we call the matrix P and take the coefficients in e^2t en te^2t together then we find

\[(P-21_{2x2})\eta=0\]
\[(P-21_{2x2})\zeta=\eta\]

The 1 of 21 in these equations are boldened. Can someone explain me where these equations come from?

Answer 2

ok

Answer 3

yup

Answer 4

It's the last two equations with P that I don't get.

Answer 5

I have time :-)

Answer 6

The course continues by saying that the first equation tells us that eta=(1,-1)^T, but there is something strange going on with the second: the determinant of the matrix equals zero, which can make us think there is no solution to the second equation (simply inverting to get a solution is not an option). For some eta's one could solve the equation: "the vectors eta that are perpendicular (Euclidian inprouct) to various freely chosen y's for the system

\[(P ^{\dagger} - 21_{2x2})y=0\]

With P^dagger the Hermitic addition. So, this all serves to claim that zeta does exist. They can be found with the system

\[-\zeta _{1} -\zeta _{2}=+1\]
\[+\zeta _{1} +\zeta _{2}=-1\]

The idea behind the above condition about eta is that to say that eta (element of R^2) must be perpendicular to vectors y in ker M^dagger is like saying that

\[R ^{2}= \ker M circledcross \ker M^\dagger\]

with

\[M = (P - 21_{2x2}\]

Answer 7

Yup the second eigenvector in the course ends up being (0,-1). But I just don't get the whole stuff about (P-21_2x2). I don't really understand its meaning and where my prof gets it from.

Answer 8

@wio

Answer 9

Why did you mention me? What is the question?

Answer 10

Are you sure it isn't \[
P - 2I_{2\times 2} = P -\begin{bmatrix}
2&0\\
0&2
\end{bmatrix}
\]

Answer 11

I need someone to explain to me what the (P-...) expressions and equations mean and where they come from.

Answer 12

It's a boldened 1, not an I

Answer 13

Well you have to look it up in your book then, because it's not standard notation.

Answer 14

I've looked all over the linear algebra course, dagger and ker I find, but not that notation. :(

Answer 15

Different people will say it means different things.

Answer 16

wikipedia will say it is identity matrix, others wills say is it vector of ones.

Answer 17

I've put the question to my professor by email, to make sure it's capital I for Identity matrix or not.

Answer 18

Thanks at least for suggesting an idea about it, @wio

Answer 19

The purpose is to get a 0 determinate, usually

Answer 20

Which will ensure the null space is a vector, your Eigen vector.

Answer 21

Yeah, I'm starting to get it now... He probably used 1 to indicate it's a vector of ones on the diagonal, but it's less clean that using identity notation.

Answer 22

the -2I_2x2 is the eigenvalue in matrix notation form.