Possible Jordan Forms for 3x3 Matrix with Negative Eigenvalues

  • Thread starter Thread starter Kamekui
  • Start date Start date
  • Tags Tags
    3x3 Forms Matrix
Kamekui
Messages
12
Reaction score
0

Homework Statement


1. Homework Statement [/b]
Enumerate all possible Jordan forms for 3 x 3 systems where all the eigen-values have negative real parts. Do not use specific values. Instead, use possibilities
like λ1; λ2; λ3, each with multiplicity 1, or λ (multiplicity 3).



Homework Equations





The Attempt at a Solution



Let Ji be the Jordan Form

J1=\begin{bmatrix}
λ1 & 0 & 0 \\
0 & λ2 & 0\\
0 & 0 & λ3
\end{bmatrix}

So λ1, λ2, and λ3 all have multiplicity 1

J2=\begin{bmatrix}
λ1 & 0 & 0 \\
0 & λ2 & 1\\
0 & 0 & λ2
\end{bmatrix}

λ1 (Multiplicity 1), λ2 (Multiplicity 2)


J3=\begin{bmatrix}
λ1 & 0 & 0\\
0 & λ1 & 0\\
0 & 0 & λ1
\end{bmatrix}

λ1 (Multiplicity 3) With 1 generalized eigenvector



J4=\begin{bmatrix}
λ1 & 1 & 0\\
0 & λ1 & 1\\
0 & 0 & λ1
\end{bmatrix}

λ1 (Mulitiplicity 3) With 2 generalized eigenvectors


J5=\begin{bmatrix}
λ1 & 0 & 0 \\
0 & λ2 & 0\\
0 & 0 & λ3
\end{bmatrix}

Where λ1 ε ℝ, λ2 and λ3 are complex conjugates such that
λ2= -a+bi and λ3=-a-bi. So λ1, λ2, and λ3 all have multiplicity 1.


1) Do these Jordan Forms look correct?
2) Are there more? ( I think there may be 3 more but I'm unsure)
 
Physics news on Phys.org
Why the blank spaces? Were those supposed to be "1"s?
 
Sorry if I seem confused but what blank spaces?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top