# Develop a fundamental matrix and the general solution?

#### kalle100

Hi,
I know how to get the first eigenvector and x1, but how do I continue when only get a lambda value which in this task becomes 1?
What should you do to get the other intrinsic value and x2?

Consider the system 1. Determine a fundamental amount of solution
2. Find the general solution

system: This is so far I have come: #### SDK

Now you find a basis of (generalized) eigenvectors.

#### romsek

Math Team
• DarnItJimImAnEngineer

#### SDK

I couldn't remember how to deal with repeated eigenvalues, but a quick Google search found this; you might try their technique.

Differential Equations - Repeated Eigenvalues
I usually think Paul's online notes do a good job of presenting calculus topics, but that page is absolutely terrible. The two most egregious problems are the following:

1. The “mystery $t$” appearing in solutions of repeated eigenvalue problems is supposed to be a short lived phase. It lasts maybe a few lectures at most. Once it is analyzed properly with linear algebra, the appearance of $t$ is completely natural. It's absurd to put supplementary notes online which not only rely on this mystery $t$ in the higher dimensional case, but actually appeal to the scalar version as a justification! This is supposed to be where things are explained properly. It's like if your kid is 16 years old and just finds out Santa isn't real and instead of agreeing you double down and insist that he is.

2. It's wrong! This is the problem with the mysterious $t$ “approach”. It just doesn't work in general. For example, $\dot x = Ix$ where $I$ is the $2\times 2$ identity matrix. This obviously has a repeated eigenvalue but the generic solutions don't include a term with a factor of $t$. More importantly, there is not an obvious reason why this shouldn't have a $t$ while other equations do unless you know why the $t$ appears.

Since this has obviously annoyed me more than expected, I wrote a quick note about this which is (in my apparently not so humble opinion) much better. Since at least 2 other people seem to think Paul's notes are a decent approach to this, I'm happy to argue about this if anyone has some justification for the “mystery $t$” approach. I'll include them in a separate post.

#### SDK

Solving linear ODEs with repeated eigenvalues correctly.

Let's start with the relationship between eigenvectors and solutions of linear ODEs. Consider the equation
$\dot x = Ax \quad x(0) = x_0 \in \mathbb{R}^n$
and assume first that $A$ is diagonalizable. That means if $\{v_1,\dotsc, v_n\}$ are the eigenvectors of $A$, then they form a basis. The solution of this equation is given simply by $x(t) = \exp(tA)x_0$ where $\exp(tA)$ is the matrix exponential defined by the formula
$\exp(tA)x_0 = \sum_{k = 0}^\infty \frac{t^k A^k}{k!}.$
I'll stick to using $\exp(*)$ for matrix exponentiation and $e^{*}$ for scalars so it is always obvious what kind of exponential is meant.

The reason we write the solution this way boils down to the following computation. Let's look at what happens if we choose $x_0$ to be an eigenvector of $A$, say $x_0 = v_i$. Then this formula says that the solution is
$x(t) = \exp(tA)v_i = \sum_{k = 0}^\infty \frac{t^k A^k}{k!}v_i = \sum_{k = 0}^\infty \frac{t^k \lambda_i^k}{k!}v_i = e^{\lambda_i t} v_i$
where $\lambda_i$ is the eigenvalue for $v_i$. This computation says that the solutions which starts in the span of an eigenvector remains in the span of that eigenvector for all $t \in \mathbb{R}$. This is what makes eigenvectors special (at least in this context, they are special for many reasons). Eigenvectors are the “straight line” solutions of the ODE.

Now, remove the assumption that $A$ has a basis of eigenvectors. Let's assume $A$ is $2 \times 2$ with repeated eigenvalue $\lambda$, but $A$ has only a single eigenvector for $\lambda$, say $v_1$. It turns out this is the simplest case. but it actually settles the entire problem for all dimensions via a trivial generalization. To rephrase the situation, we have $\operatorname{ker}(A - \lambda I) = \operatorname{span}(\{v_1\})$ which raises the important question, where does $A - \lambda I$ map the rest of the vectors in $\mathbb{R}^2$?

It turns out (a consequence of the Cayley-Hamilton theorem) that in this case there must exist a $v_2 \notin \operatorname{span}(\{v_1\})$ which satisfies the equation $\left(A - \lambda I\right)^2 v_2 = 0$. But that means the vector $\left(A - \lambda I\right)v_2 \in \operatorname{ker}(A) = \operatorname{span}(\{v_1\})$ so after rescaling appropriately, we can simply assume that $v_1 = \left(A - \lambda I\right)v_2$. The vector, $v_2$ is called a generalized eigenvector for $A$. Now, let's look at what this means for the ODE.

To start, let's revisit the computation of the solution for an eigenvector. Only this time we rewrite the condition of being an eigenvector in the form $\left(A - \lambda I\right) v_1 = 0$ so then taking the matrix exponential, the same computation is equivalent to
$\exp\left(t \left(A - \lambda I\right)\right) v_1 = \sum_{k = 0}^\infty \frac{t^k \left(A - \lambda I\right)^k}{k!} v_1 = v_1$
where we used the fact that $\left(A - \lambda I\right)^kv_1 = 0$ for all $k \geq 1$. It turns out that if two matrices, $A$ and $B$, commute with one another, then the matrix exponential satisfies the nice property as in the scalar case: $\exp{AB} = \exp{A}\exp{B}$. Since the identity commutes with any matrix, we apply this one the left of the previous equation to get
$\exp\left(t \left(A - \lambda I\right)\right) v_1 = \exp \left(tA\right) \exp \left(-\lambda t I\right)v_1 = e^{-\lambda t} \exp \left(t A\right)v_1.$
Combining these equations, we arrive at the identity
$e^{-\lambda t} \exp \left(t A\right)v_1 = v_1 \implies \exp \left(tA\right)v_1 = e^{\lambda t} v_1$
which is just the same identity we already knew.

However, doing the same computation for $v_2$ now reveals where the mysterious $t$ arises from. Start with
$\exp\left(t \left(A - \lambda I\right)\right) v_2 = \sum_{k = 0}^\infty \frac{t^k \left(A - \lambda I\right)^k}{k!} v_2 = v_2 + t \left(A - \lambda I\right)v_2$
where the additional term comes from the fact that now, $\left(A - \lambda I\right)^kv_2 = 0$ only if $k \geq 2$. Applying our nice multiplicative identity again gives us
$e^{-\lambda t} \exp \left(t A\right)v_2 = v_2 + t \left(A - \lambda I\right)v_2 \implies \exp \left(tA\right)v_2 = e^{\lambda t} v_2 + t e^{\lambda t}\left(A - \lambda I\right)v_2.$
But remember that $\left(A - \lambda I\right)v_2 = v_1$, so we end up with the solution
$\exp(tA)v_2 = e^{\lambda t} v_2 + t e^{\lambda t} v_1.$

#### romsek

Math Team
Since at least 2 other people seem to think Paul's notes are a decent approach to this I'm happy to argue about this if anyone has some justification for the mystery $t$'' approach. I'll include them in a separate post.
I wouldn't say I think Paul's notes are a decent approach. I didn't read through section. It seemed to address OP's issue and I left it to OP to decide whether it satisfied their needs.
If not they could come back and ask more. The likelihood of just being ignored is just too high to devote a ton of time to an answer w/o feedback from OP.