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Find the general solution of $X'=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}X$

What I did: I calculated that the matrix has repeated eigenvalues of 0. When I plugged this in, I got that an eigenvector is $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. So one solution of the system is $c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ for a constant $c_1$. But I can't find another solution.

cppcoder
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1 Answers1

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This one is easier to solve directly, since the system reduces to $x' = y$ and $y' = 0$. I would say eigenvalues and so on would be overkill here.

Zarrax
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  • Yes, I have got it now. I also got the same result using the method in the link @Gregory sent. Thanks! – cppcoder Apr 11 '20 at 22:23
  • Yes it is quite clear that this equation is simple to solve without the method I linked to in comments, but it seemed that the OP was meant to get some familiarity with how solving that type of system would work. – Gregory Apr 12 '20 at 00:27