We have two eigenvalues $\lambda_1$ and $\lambda_2$, with $\lambda_1 = \lambda_2 < 0$. Consider the case we have only one linearly independent eigenvector. Then the phase portrait is a (stable) degenerate node. I have seen two different phase portraits for this case:
In general, how we decide which phase portrait we need?
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$\begingroup$When the $2\times2$ 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 sign of $\det B$ determines the “handedness” of the phase portrait: with a positive determinant, the phase portrait will be similar to the left-hand one pictured. 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).
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