I have a question with respect to phase plots of repeated eigenvalue cases. For instance suppose that one is given a matrix with the following:
$$\overrightarrow{y'} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \overrightarrow{y}$$
whereby the solution yields:
$$\overrightarrow{y} = c_1e^t\binom{2}1 + c_2(te^t\binom{2}1 + e^t\binom{1}0)$$
I understand that it is asymptotically unstable but what I am unclear is the direction and whether one should plot it according to the following two figures
How does one distinguish which way the curves come out of the origin from just from analyzing the system ? I am aware that the plot should actually be the 2nd picture but I am not sure why so some help with respect to this technique would be appreciated.
When the $2×2$ matrix $A$ has a repeated eigenvalue $\lambda$ (with $A\ne \lambda I$), we have $$A=B\pmatrix{\lambda&1\\0&\lambda}B^{−1}$$ and $$\exp(tA)=Be^{\lambda t}\pmatrix{1&t\\0&1}B^{−1}.$$ The sign of $\lambda$ determines whether the flow is inward or outward, while the signs of $\lambda$ and $\det B$ determine the “handedness” of the phase portrait: if the signs are the same, the phase portrait will be similar to the first one pictured; if they are different, it will be like the second. You can see this for yourself by comparing the simple cases $B=I$ and $B=\pmatrix{−1&0\\0&1}$ (the eigenvector lies along the $x$-axis in both cases) using positive and negative values for $\lambda$.
In your case, $\lambda=1$, $B=\pmatrix{2&1\\1&0}$ and $\det B=-1$, so you’ll get the second phase portrait with outward flow.