# How Are Generalized Eigenvalues used in Differential Equations?

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?

• What is your level of math and is this course lower or upper division? This will help in what I can use answering your question.
– help
Nov 14, 2019 at 5:07
• This weekend I answered a related question that might help: math.stackexchange.com/a/3429450/596135 Nov 14, 2019 at 5:10
• @BlackBlast: See Theorem 5: uio.no/studier/emner/matnat/math/MAT2440/v11/…. I also got different eigenvectors than you wrote.
– Moo
Nov 14, 2019 at 5:36
• The main thrust of the argument is that you want to transform to a basis where your differential equations look "simple" or rather "as simple as possible". If the matrix were diagonalizable, then all of the variables can be decoupled and you simply have to transform back to your original basis. If the matrix can't be diagonalized, then it will at least have a Jordan canonical form with a block diagonal. You can still solve the ODE by back substitution and transform back afterwards. Nov 14, 2019 at 5:57

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.