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Let's imagine the following scenario:

  • algebraic multiplicity: $\lambda_{1} = \lambda_{2} = \lambda_3: 3$
  • geometric multiplicity: 1

the first column of the fundamental matrix can be found as follows using row expansion:

$\vec{y_{1}(t)} = e^{\lambda_{1}t}[I\vec{x_{1}}+ \frac{t^1}{1!}(A-\lambda_{1}I)\vec{x_{1}}+...+\frac{t^n}{n!}{(A-\lambda_{1}I)}\vec{x_{1}}] = e^{\lambda_{1}t}.\vec{x_{1}} $

Now we need to create a new vector ourselves, as we only have one eigenvector:

$ \vec{y_{2}(t)} = e^{\lambda_{1}t}[I\vec{x_{2}}+ \frac{t^1}{1!}(A-\lambda_{2}I)\vec{x_{2}}+...+\frac{t^n}{n!}{(A-\lambda_{2}I)}\vec{x_{2}}] = e^{\lambda_{2}t}(t\vec{x_{1}}+\vec{x_{2}}) $

where $\frac{t^1}{1!}(A-\lambda_{2}I)\vec{x_{2}} = \vec{x_{1}}$ and all the other terms become $0$.

And let's now create the third and final vector:

$ \vec{y_{3}(t)} = e^{\lambda_{3}t}[I\vec{x_{3}}+ \frac{t^1}{1!}(A-\lambda_{3}I)\vec{x_{3}}+...+\frac{t^n}{n!}{(A-\lambda_{3}I)}\vec{x_{3}}] = e^{\lambda_{3}t}(\vec{x_{3}}+t\vec{x_{2}}+ \frac{t^2}{2!}\vec{x_{1}}) $

with $(A-\lambda{3}I)\vec{x_{3}} = \vec{x_{2}}$ and all the other terms equal $0$.

My question is: are the columns of the fundamental matrix which is composed of those three vectors (2+1), still linear independent? The two last vectors are calculated using the first eigenvector. This means, I think, that the three vectors won't be linear independent, meaning the columns of the fundamental matrix will not be linear independent neither.

But when solving a system of first order differential equations, converted to matrices, using the eigenvalues and eigenvectors the solutions are in the columns of the FM. And they have to be linear independent. Isn't that paradoxal?

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If you've learned about Jordan canonical form, you'll recognize that $x_1,x_2,x_3$ give a basis that puts the matrix in Jordan form. The vectors are indeed linearly independent. Have you tried working out examples?

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  • $\begingroup$ Since you asked about examples, I will mention that I talked with the OP a bit in chat. The discussion involved the simplest example I was able to think of. (And the main purpose was to try to understand they want to ask.) $\endgroup$ – Martin Sleziak Jan 24 '16 at 8:58
  • $\begingroup$ Thanks, @MartinSleziak. If the OP pursues it, I'll add a $2\times 2$ example. $\endgroup$ – Ted Shifrin Jan 24 '16 at 17:10

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