Hi all,

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I have a test tommorow, and I'm currently learing how to solve complex systems of ODEs, and I'm stuck on finding the eigenvectors once I found the eigenvalues. Basically, I don't understand why you need to row reduce and make the second row equal to zero, even if you can already infer what the eigenvector is going to look like from one of the rows, for example when there is a 1.

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As an example, take the system x'=(2, -5 ; 1, -2)x. the eigenvalues are i and -i. So I set up ( 2-i, -5 ; 1, -2-i )( v_1 ; v_2 )=( 0 ; 0 ). Since the bottom row is equivalent to v_1 = (2 i)*v_2, why can't you just use that to say that that the eigenvector for the eigenvalue i is ( 2 i ; 1 )? If I row reduce, I get ( 1, -5/(2-i) ; 0, 0 )( v_1 ; v_2 )=( 0 ; 0 ), so the eigenvector for i now changes to ( 5/(2-i) ; 1 ).

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Another example is for x'=(1, -1 ; 5, -3)x where the eigenvalues are -1 i and -1-i. For -1-i, the system is ( 2 i, -1 ; 5, -2 i )( v_1 ; v_2 )=( 0 ; 0 ), and so I though you could just say that since the top row implies that (2 i)*v_1 = v_2, the eigenvector is ( 1 ; 2 i ). But once you row reduce to get ( 1, -(2 i)/5 ; 0, 0 )( v_1 ; v_2 )=( 0 ; 0 ), so the eigenvector is ( (2 i)/5 ; 1 ).

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This completely changes the answer, so I'm very confused. The first example in particular, because I solved it without row reducing first, and my answer matched the one from symbolab, but not my textbook solution. Are these just scalar multiple maybe? and both answers are correct? Or do I need to row reduce to get the correct answer? If that is the case, why does the relationship I find without row reducing now hold?

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Thanks