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I have a dynamical system: $\dot{\mathbf x}$= A$\mathbf x$ with $\mathbf x$= $\bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ and A = $\bigl( \begin{smallmatrix} 3 & 0 \\ \beta & 3 \end{smallmatrix} \bigr). \beta$ real, time-independent.

I calculated the eigenvalue $\lambda$ = 3 with the algebraic multiplicity of 2.

The first question is about eigenvectors when $\beta = 0$ and when $\beta \neq$ 0:

1) when $\beta$ = 0, I have $\bigl( \begin{smallmatrix} 0 & 0 \\ \beta & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 \\ 0\end{smallmatrix} \bigr)$. Does this allow for eigenvectors calculation? Does it tell me anything at all?

2) when $\beta \neq$ 0, I have $\bigl( \begin{smallmatrix} 0 & 0 \\ \beta & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 \\ 0\end{smallmatrix} \bigr)$, so my eigenvector is $\bigl( \begin{smallmatrix} 0 \\ 1\end{smallmatrix} \bigr)$ and A is defective? Is there any other eigenvector?

The other question is further on case 2) when $\beta$ = 3. I am to find any fixed points + their stability, but first I am wondering whether I did the above correctly. I am not sure how to approach it, thought about a trajectory expression, but I am confused by the defective A.

Thanks!

Carlos
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Donnie D
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3 Answers3

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Although it does not really matter, it is traditional (well, in the U.S.) to put the $1$ in the Jordan form above the diagonal. I have been noticing students lately getting to the Jordan form but failing to write things in the reverse( and actually useful) order. $$ \left( \begin{array}{rr} 0 & \frac{1}{\beta} \\ 1 & 0 \end{array} \right) \left( \begin{array}{rr} 3 & 0 \\ \beta & 3 \end{array} \right) \left( \begin{array}{rr} 0 & 1 \\ \beta & 0 \end{array} \right) = \left( \begin{array}{rr} 3 & 1 \\ 0 & 3 \end{array} \right) $$

$$ \left( \begin{array}{rr} 0 & 1 \\ \beta & 0 \end{array} \right) \left( \begin{array}{rr} 3 & 1 \\ 0 & 3 \end{array} \right) \left( \begin{array}{rr} 0 & \frac{1}{\beta} \\ 1 & 0 \end{array} \right) = \left( \begin{array}{rr} 3 & 0 \\ \beta & 3 \end{array} \right) $$

Then the exponential of $Jt$ with $$ J = \left( \begin{array}{rr} 3 & 1 \\ 0 & 3 \end{array} \right) $$ is $$ e^{3t} \; \; \left( \begin{array}{rr} 1 & t \\ 0 & 1 \end{array} \right) $$ and you use the second matrix identity above to finish

$$ e^{At} = e^{3t} \left( \begin{array}{rr} 0 & 1 \\ \beta & 0 \end{array} \right) \left( \begin{array}{rr} 1 & t \\ 0 & 1 \end{array} \right) \left( \begin{array}{rr} 0 & \frac{1}{\beta} \\ 1 & 0 \end{array} \right) = e^{3t} \left( \begin{array}{rr} 1 & 0 \\ \beta t & 1 \end{array} \right) = \left( \begin{array}{rr} e^{3t} & 0 \\ \beta t e^{3t}& e^{3t} \end{array} \right) $$

Will Jagy
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When $\beta = 0$, you have $A = 3I$ so the eigenvectors are the standard basis and the eigenvalues are both $1$.

When $\beta \ne 0$, the matrix is not diagonalizable, it only has a Jordan form.

gt6989b
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The eigenvectors for $\beta\ne0$ are $[0,1]^T$, linear dependent, therefore they don’t form a basis (not diagonalizable), thus defective.

The eigenvectors for $\beta=0$ are $[0,1]^T$, $[1,0]^T$, i.e. ,the standard basis.

The fixed points do not change for $\beta\ne0$, and they are clearly unstable since the eigenvalues are positive.

Carlos
  • 691