Homogenous Linear Systems of Equations

**Fwahm** · February 22nd 2010, 03:07 PM

I have this problem, and could only get part way through before I got stuck. Here it is.

Solve the given system subject to the initial condition.

X'=|2 4| X, X(0) = |-1|
---|-1 6| --------- |6 |

I already solved for lambda (4 and 4), and their eigenvectors (2,1 and 2,1), but I'm stuck after this point. Any help?

**HallsofIvy** · February 23rd 2010, 05:59 AM

Originally Posted by Fwahm

I have this problem, and could only get part way through before I got stuck. Here it is.

Solve the given system subject to the initial condition.

X'=|2 4| X, X(0) = |-1|
---|-1 6| --------- |6 |

I already solved for lambda (4 and 4), and their eigenvectors (2,1 and 2,1), but I'm stuck after this point. Any help?

I started to say this was easy until I noticed that you have a double eigenvalue. That makes it harder. I'm not sure what you mean by saying "their eigenvectors are (2,1) and (2,1)". That is, of course, a single eigenvector which means that there is not a "complete set" of eigenvectors- there is not a basis for $R^2$ consisting of eigenvectors for this matrix and so we cannot diagonalize it.

What we can do is find a "generalized eigenvector" to convert to Jordan Normal form.

Since 4 is a double eigenvalue, the characteristic equation is $(\lambda- 4)^2= 0$ . Since every matrix satisfies its own characteristic equation, we must have $(A- 4I)^2= 0$ or $(A- 4I)^2v= 0$ for every v in $R^2$ . Obviously, if v is an eigenvector, then (A-4I)v= 0 so $(A-4I)^2=0$ follows.

But since the only eigenvector is (2,1) and that does not span $R^2$ , we must have vectors, v, such that $(A-4I)v\ne 0$ but $(A-4I)^2v= 0$ . Well, if we let u= (A- 4I)v, then $(A-4I)^2v= (A- 4I)u= 0$ so u= (A- 4I)v must be a multiple of (2, 1). In fact, we can take that "multiple" to be 1 and look for v such that (A- 4I)v= (1, 2). That is $\begin{bmatrix}-2 & 4 \\ -1 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}2 \\ 1\end{bmatrix}= 0$ which reduces to the two equations -2x+ 4y= 2 and -x+ 2y= 1. Dividing the first equation by 2 shows that it is the same as the second equation: there are an infinite number of solutions (which must be true in order that we be able to span $R^2$ ) but, taking x= 1, one is (1, 1). This is a "generalized eigenvector".

Form the matrix, P, having those vectors as columns: $P= \begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}$ . It is easy then to calculate that $P^{-1}= \begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}$ .

The point of all of that is that $P^{-1}AP=$ $\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}\begin{bmatrix}2 & 4 \\ -1 & 6\end{bmatrix}$ $\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}$ $= \begin{bmatrix} 4 & 1 \\ 0 & 4\end{bmatrix}$ , the "Jordan Normal Form" and as close to a diagonal form as we can get.

Now, the original equation is of the form $X'= AX= A(PP^{-1})X$ . Multiply on both sides of the equation by $P^{-1}$ and we have $P^{-1}X'= (P^{-1}AP)(P^-1X)$ . If we let $Y= P^{-1}X$ , our equation becomes $Y'= \begin{bmatrix}4 & 1 \\ 0 & 4\end{bmatrix}Y$ . If we write $Y= \begin{bmatrix} x \\ y\end{bmatrix}$ , that becomes $\begin{bmatrix}x' \\ y'\end{bmatrix}$ $= \begin{bmatrix}4 & 1\\0 & 4\end{bmatrix}$ $\begin{bmatrix} x \\ y\end{bmatrix}$ .

We can write that the two equations x'= 4x+ y and y'= 4y. Those are not "completely uncoupled" as would have happened if we had been able to "diagonalize" the matrix (if there had be two independent eigenvectors) but they are "partially uncoupled". We can solve y'= 4y immediately: $y(t)= Ce^{4t}$ .

Now we put that into the second equation: $x'= 4x+ Ce^{-4t}$ . Solving the corresponding homogeneous equation, x'= 4x, gives $x= De^{4t}$ again. And since the non-homogeneous part is also of that form we try a "specific solution" of the form $x= Ate^{4t}$ . Then $x'= Ae^{4t}+ 4Ate^{4t}$ and the equation becomes $Ae^{4t}+ 4Ate^{4t}= Ae^{4t}+ Ce^{-4t}$ . The two $Ae^{4t}$ terms cancel and we must have A= C/4. $x(t)= De^{4t}+ (C/4)te^{4t}$ .

That means that $Y(t)=\begin{bmatrix}De^{4t}+ (C/4)te^{4t} \\ Ce^{4t}\end{bmatrix}$ .

I'll let you do the last step. Since we defined $Y= P^{-1}X$ , $X= PY= \begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}$ $\begin{bmatrix}De^{4t}+ (C/4)te^{4t} \\ Ce^{4t}\end{bmatrix}$ .

Once you have it back to X, you can use the initial values to find C and D.

There are some "shortcuts" you could use but I did not know what you had done before so I gave you the "whole thing".

Thread: Homogenous Linear Systems of Equations

LinkBack

Thread Tools

Display

Homogenous Linear Systems of Equations

Similar Math Help Forum Discussions

Homogenous system of linear equations.

System of second order linear homogenous differential coupled equations

Non-Homogenous 2nd Order Linear Equations

General solutionsof linear second order homogenous diff equations

2nd order homogenous in to 1st order linear systems

Search Tags