Consider $X' = AX$ where $A = \begin{pmatrix} a & 1 \\ 2a & 2 \end{pmatrix}$. For which values of $a$ do you find a bifurcation?
I attempted to solve this by finding the eigenvalues which I got $\lambda = 0, 2+a$ and corresponding eigen vectors $\begin{pmatrix} 1 \\ -a \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \end{pmatrix}$. When I put these in canonical form I end up with $\begin{pmatrix} 0 & 0 \\ 0 & 2+a \end{pmatrix}$.
So I would think the only bifurcation point would be $a = -2$ because that causes the matrix to turn into the zero matrix. Otherwise, for any other value of $a$ you would essential have the same canonical form of the matrix. However, I plugged these equations into pplane graphing software and it seems as though there may also be a bifurcation point at $a = 0$ because the behavior is completely different. Is this because $0$ is another eigen value in the canonical form matrix? Are there any other bifurcation points that I should be aware of?
We are given $X' = AX$ where:
$$A = \begin{pmatrix} a & 1 \\ 2a & 2 \end{pmatrix}$$
We are asked to find the values of $a$ that produce a bifurcation.
The eigenvalues/eigenvectors are given by:
$$\lambda_1 = 0, v_1 = \begin{pmatrix} 1 \\ -a \end{pmatrix}, \lambda_2 = a + 2, v_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
The first eigenvalue is fixed as $\lambda_1 = 0$, so lets look at the second one.
For $\lambda_2 = a + 2$, we have three cases:
This tells us that there is a change in stability at $a = -2$ and that is our bifurcation point.
If we write the solution for this system, we have:
$$X(t) = c_1 \begin{pmatrix} 1 \\ -a \end{pmatrix} + c_2 e^{(a+2)t} \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
Notice what happens to this solution at $a \lt -2, a = -2, a \gt -2$?
If we were to draw phase portrait movie by varying $a$ between $a \in (-3,3)$ in steps of $0.5$, we can see the stability transition (look at the direction arrows).
Note: at $a = 0$, we have nothing special.