While solving non linear system I got this matrix and don't know how to get the eigenvectors to draw the phase portrait. I got zero for both eigenvectors. \begin{bmatrix} 3 & 0 \\ 0& 2 \end{bmatrix} could you help explaining this?
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Both are non-zero, check the determinant. – 19aksh Jun 12 '19 at 10:41
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1How do you compute eigenvalues? – Bernard Jun 12 '19 at 10:42
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1The eigenvalues of a diagonal matrix are the diagonal elements... – StackTD Jun 12 '19 at 10:47
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1I meant I got zero for egenvectors not values – F.O Jun 12 '19 at 10:49
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Ahaha, I see what happened.
Your first equation for, let's say $\lambda=3$, would be $0x+0y=0$ (meaningless) and $0x-1y=0$, Then obviously, you have $y=0$. However, since it's $0x$, really $x$ can be anything. Which means the eigenvector is $\begin{pmatrix}n\\0 \end{pmatrix}$. Or we just norm it as $\begin{pmatrix}1\\0 \end{pmatrix}$.
Similarly you get $\begin{pmatrix}0\\1 \end{pmatrix}$ for $\lambda=2$.
Saketh Malyala
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