# Diff eqs with eigenvectors: double roots, but 2nd eigenvector?

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1. Nov 15, 2015

### kostoglotov

The problem is here, I'm trying to solve (b):

and the text solution is here:

I understand why there is a term in there with $cte^t$, it's because the A matrix has double roots for the eigenvalues. What I don't understand is where the (apparent) second eigenvector, $\begin{bmatrix}1\\ t\end{bmatrix}$ is coming from?

I gave my answer as $\vec{u} = \begin{bmatrix}4\\ 2\end{bmatrix} + c_1e^t\begin{bmatrix}0\\ 1\end{bmatrix}+c_2te^t\begin{bmatrix}0\\ 1\end{bmatrix}$.

This answer works, but so does the text answer, and it is more complete. But where did that second distinct eigenvector come from?

2. Nov 15, 2015

### Staff: Mentor

It comes from $A_{21} = 1$ and the exponential function but I don't know how to put it correctly. Too long ago.

3. Nov 16, 2015

### Krylov

You have a double eigenvalue $\lambda = 1$ with geometric multiplicity one, giving you one eigenvector $v_1 = (0,1)$, and algebraic multiplicity two. So, in order to span the generalised eigenspace corresponding to $\lambda$ (which is just $\mathbb{R}^2$) you need a generalised eigenvector $v_2$, which you can obtain by solving
$$Av_2 = \lambda v_2 + v_1$$
(The sequence $\{v_1,v_2\}$ is called a Jordan chain corresponding to $\lambda$ and $v_1$.) Then the solution to the homogeneous system is
$$u_n(t) = c_1 e^{\lambda t}v_1 + c_2 t e^{\lambda t}v_2$$
You can see that your own solution is not the most general one, because at time $t = 0$ you cannot satisfy an arbitrary initial condition. (In fact, as you can see you can only satisfy initial conditions $u_0$ for which $u_0 - (4,2)$ is in the span of $(0,1)$.) This is because $v_1$ by itself does not span the generalised eigenspace.

4. Nov 16, 2015

### Staff: Mentor

A different approach from the one Krylov took...
Since it is straightforward to get a particular solution to the nonhomogeneous problem, let's look only at the homogeneous system:
$\vec{u}' = A\vec{u}$
This system has a solution $\vec{u} = e^{At}C$, where C is a column matrix of coefficients that depend on initial conditions.

To calculate $e^{At}$, we'll need to calculate the various powers of A.
$A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$
$A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$
$A^3 = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$
In general, $A^n = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$
If necessary, this last statement is easy to prove.

$e^{At} = I + At + \frac{A^2t^2}{2!} + \dots + \frac{A^nt^n}{n!} + \dots$
$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + t\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \frac{t^2}{2!}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} + \dots + \frac{t^n}{n!}\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix} + \dots$
$= \begin{bmatrix} 1 + t + \frac{t^2}{2!} + \dots + \frac{t^n}{n!} + \dots & 0 \\ 0 + t + \frac{2t^2}{2!} + \dots + \frac{nt^n}{n!} + \dots & 1 + t + \frac{t^2}{2!} + \dots + \frac{t^n}{n!} + \dots \end{bmatrix}$
$= \begin{bmatrix} e^t & 0 \\ te^t & e^t \end{bmatrix}$
$= e^t\begin{bmatrix} 1 & 0 \\ t & 1 \end{bmatrix}$
In the matrix a couple of lines up, the expression in the lower left corner is just tet.

So the solution to the homogeneous problem, $u_h(t)$, is $u_h(t) = e^t\begin{bmatrix} 1 & 0 \\ t & 1 \end{bmatrix}C$,
or, $u_h(t) = c_1e^t\begin{bmatrix} 1 \\ t \end{bmatrix} + c_2e^t\begin{bmatrix} 0 \\ 1 \end{bmatrix}$
In this last form, the two linearly independent eigenvectors are shown.

5. Nov 17, 2015

### Krylov

Mark44 made me realise that I made a mistake in this line yesterday, my apologies. The method is fine, though. From the Jordan chain you build the correct solution to the homogeneous equation as
$$u_n(t) = c_1e^{\lambda t}v_1 + c_2e^{\lambda t}(v_2 + t v_1)$$
I'm sorry for any confusion that I may have caused.

6. Nov 17, 2015

### Krylov

To make you see the pattern of constructing solutions using Jordan chains more clearly, I will add one example. Suppose you have the homogeneous $3 \times 3$ system corresponding to the matrix
$$A = \begin{bmatrix} 1& 0& 0\\ 1& 1& 0\\ 1& 1& 1 \end{bmatrix}$$
You can check that $\lambda = 1$ is a triple eigenvalue of geometric multiplicity one. The lone eigenvector is given by $v_1 = (0,0,1)$. To get the first generalised eigenvector, solve
$$A v_2 = \lambda v_2 + v_1$$
to obtain $v_2 = (0,1,0)$. Similarly, to get the second generalised eigenvector, solve
$$A v_3 = \lambda v_3 + v_2$$
to obtain $v_3 = (1, -1, 0)$. (There is not really a need for Gaussian elimination for these systems: you can find the solutions by inspection.) Then construct the solution as
$$u_h(t) = c_1 e^{\lambda t} v_1 + c_2 e^{\lambda t}\Bigl(v_2 + \frac{t}{1!}v_1\Bigr) + c_3 e^{\lambda t}\Bigl(v_3 + \frac{t}{1!} v_2 + \frac{t^2}{2!}v_1\Bigr)$$
where the $c_i$ are constants determined by your initial condition.

In my opinion, taking the matrix exponential directly (as in @Mark44 's post) is faster when it is easy to spot a general formula for the powers of $A$. This is not always the case, in which instance you can rely on computing Jordan chains.

7. Nov 17, 2015

### Staff: Mentor

In addition to the technique of calculating $e^{At}$ using powers of A, there's another technique that uses the Cayley-Hamilton theorem, which says that every matrix is a root of its characteristic equation. This technique appears in a Linear Algebra textbook I've hung onto, "Linear Algebra and Differential Equations," by Charles G. Cullen. Cullen's explanation isn't clear to me, so when I figure out what he's doing, I'll post an explanation using that technique.

8. Nov 17, 2015

### kostoglotov

But by inspection you can see that there are multiple solutions to those equations for the generalized eigenvectors. $(0,1,1)$ would work just as well to solve $(A-\lambda I)v_2 = v_1$.

9. Nov 17, 2015

### Staff: Mentor

So here's the other method I mentioned in my previous post. The technique is presented in "Linear Algebra and Differential Equations," by Charles G. Cullen.

We're solving the DE $\frac{d \vec{u}}{dt} = A \vec{u}$, or $(D - A)\vec{u} = 0$, for which the solution is $\vec{u} = e^{At}C$, where A and C are matrices.
For the problem at hand, $A = \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}$

As was already mentioned in this thread, the matrix A has only one eigenvalue: $\lambda = 1$.
The characteristic polynomial for the matrix is $c(x) = (x = 1)^2 = x^2 - 2x + 1$, which is found by evaluating det($\lambda$I - A).

For a diff. equation $(D - 1)^2y = 0$, we would expect a solution of the form $y = c_1 e^t + c_2te^t$.

Per a theorem by Ziebur, cited on page 307 of this textbook, "Every entry of $e^{At}$ is a solution of the nth-order equation c(D)y = 0, where c(x) = det(xI - A) is the characteristic polynomial of A."

That is, $e^{At} = E_1e^t + E_2te^t$, where $E_1$ and $E_2$ are matrices of constants.

At t = 0, the equation above results in $I = E_1$
Differentiating the equation above results in $Ae^{At} = E_1e^t + E_2e^t + E_2te^t$
At t = 0, we have $A = E_1 + E2$

Substituting I for $E_1$ and solving the second equation, we have $E_2 = A - I = \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}$

Therefore, $e^{At} = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}e^t + \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}te^t$
$= \begin{bmatrix} 1 & 0 \\ t & 1\end{bmatrix}e^t = e^{At}$
Note that the columns of this last matrix hold the two linearly independent eigenvectors.

Last edited: Dec 24, 2015
10. Nov 17, 2015

### Krylov

True, you could equally well work with that choice for $v_2$, but for the same initial condition $u_0$ as before the arbitrary coefficients $c_i$ would then be different and you would end up with exactly the same solution $u_h$ satisfying $u_h(0) = u_0$.

(Note that once you make a different choice for $v_2$, your solution for $v_3$ will also change.)

Last edited: Nov 17, 2015