How Are Generalized Eigenvalues used in Differential Equations?

Question

I understand the process for how Eigenvalues are involved in Differential Equations. If you have Differential System of Equations like this

$$ \vec{x}' = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}\vec{x} $$

The solution to that System of Differential Equations is a Linear Combination of e to the power of the eigenvalues times the corresponding eigenvectors.

$$ \vec{x} = C_1e^{2t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + C_2e^t\begin{pmatrix} -1 \\ 1 \end{pmatrix} $$

However, what I am struggling with figuring out how Generalized Eigenvalues translate to the solutions in Differential Equations. If you take this System of Differential Equations

$$ \vec{x}' = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \end{pmatrix}\vec{x} $$

The solution to this Differential Equations is

$$ \vec{x} = C_1e^t\begin{pmatrix} 1 \\ 3t \\ 6t + \frac{9}{2}t^2 \end{pmatrix} + C_2e^t\begin{pmatrix} 0 \\ 1 \\ 3t \end{pmatrix} + C_3e^t\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$

But the eigenvectors of this matrix (including the generalized eigenvectors) are

$$ \vec{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix},\:\: \vec{w}_1 = \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix},\:\: \vec{w}_2 = \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} $$

These eigenvectors do share some similarities to the solution to the System of Equations. However, the $$ 6t + \frac{9}{2}t^2$$ term in the solution is one I have no idea how generalized eigenvectors relate. May someone please explain how these eigenvectors translate into Differential Equations?

Answer 1

In a broad sense, the first $A=2\times 2$ matrix you wrote has unique eigenvalues of $\lambda_1=1$ and $\lambda_2=2$. This means there are two eigenvectors $x_1,x_2$ corresponding to those eigenvalues which we understand as $Ax_1=\lambda_1 x_1$ and $Ax_2=\lambda_2 x_2$. These eigenvectors will be linearly independent which you can check. This means that those two eigenvectors form a basis for a vector space, which is important because this vector space is called the solution space for your differential system. So, all the solutions to the differential system you have written, live within this certain vector space.

The second matrix you have written ($3\times 3$) has only one eigenvalue that is repeated three times. Therefore, you will not get three linearly independent eigenvectors. In fact, you get only one eigenvector in this case. This is where you have to extend from: $x_1$ for the first eigenvector, then $tx_1+x_2$ for the second, and $\frac{t^2}{2}x_1+tx_2+x_3$ as the 3rd eigenvector. This process is the way to generate a linearly independent set of vectors to form a basis for your solution space using the parameter $t$. It is not the same space, but it holds similar properties to the a vector space which would be generated by unique eigenvalues. This is the process of generalized eigenvectors.

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Typically, the question you're asking can be answered very rigorously, by defining a vector space, an element from the space of all linear functionals over that space, and using matrix representation. We would want to check algebraic multiplicity and possibly extend the set of linearly independent vectors to form a complete basis for the original space. Normally, the generalized eigenvectors on their own form a basis for an invariant subspace, only extending them forms a basis for a full vector space. The solutions to any differential equation live in the kernel of the space, which may or may not have meaning to you depending on your level of math. The more linear algebra you take, and all the upper division differential equations courses answer more of these questions you have.