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When solving the problem below:

Problem

I find that there is one eigenvalue equal to 0 and an eigenvector equal to [1; 0; 1]. How do you go about solving this system so the solution isn't just constants?

Brandon Iles
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  • I understand the meaning of algebraic and geometric multiplicity but a little confused about generalized eigenvectors – Brandon Iles Apr 16 '18 at 20:28
  • Hmmm the generalized eigenvectors now make sense but I'm getting a different answer than you. I get: x(t)=c1(1/2)(t^2), y(t)=(c2t)+c3, and z(t) = (c1(1/2)(t^2)) + (c2t) + (2*c3) – Brandon Iles Apr 16 '18 at 21:05
  • Since one eigenvector is 0, then it has multiplicity three. The first eigenvector is v1=[1;0;1], the next generalized is v2=[0;1;1], and the third is v3=[0;1;2]. Then the general solution would be x(t) = (c1(0.5(t^2)))v1 + c2tv2 + c3v3. Or x(t)=c1(1/2)(t^2), y(t)=(c2t)+c3, and z(t) = (c1(1/2)(t^2)) + (c2t) + (2*c3) – Brandon Iles Apr 16 '18 at 21:13

1 Answers1

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The solution is of course $$ X(t) = \exp(t A) X(0) $$ and as $A^3=0$ this series reduces to simply $$ X(t) = (I + tA + \tfrac12t^2A^2) X(0) $$

Lutz Lehmann
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  • That's a good point, if we already have nilpotent we do not need Jordan form; I came across something similar I had not known: https://math.stackexchange.com/questions/2722246/finding-nilpotent-and-diagonalizable-operators-for-a-linear-transformation-t/2722317#2722317 – Will Jagy Apr 16 '18 at 21:39