Finding Eigenvectors for a Homogenous System with Three Lambda Values

[cue928]

I am working on problems involving the eigenvalue method for homogenous systems in my differential equations class. However, I have not had linear algebra yet so I am not understanding some of this. I've made it to the step where I have three lambda values and plugging them back into the matrix with the identity, I have the following matrix:

3 1 3
1 5 1
3 1 3

How do I calculate the eigenvectors for this matrix? I mean, I have no clue how to do this. I was looking at a few problems in the solution manual and it appears some could be handled thru inspection. But, I didn't see a way to do that on this one.

 

[vela]

The first and third row are identical, so you will get only two independent equations from that matrix:

\begin{align*}<br /> 3x + y + 3z &amp;= 0 \\<br /> x + 5y + z &amp;= 0<br /> \end{align*}

Because you have three unknowns but only two equations, you can solve for any two of the variables in terms of the third. In this case, for example, you'll find x=-z and y=0, so you can write

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -z \\ 0 \\ z \end{pmatrix} = z\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}

You can arbitrarily set z=1, so the eigenvector is (-1, 0, 1).

 

[LCKurtz]

I am working on problems involving the eigenvalue method for homogenous systems in my differential equations class. However, I have not had linear algebra yet so I am not understanding some of this. I've made it to the step where I have three lambda values and plugging them back into the matrix with the identity, I have the following matrix:

3 1 3
1 5 1
3 1 3

How do I calculate the eigenvectors for this matrix? I mean, I have no clue how to do this. I was looking at a few problems in the solution manual and it appears some could be handled thru inspection. But, I didn't see a way to do that on this one.

You use elementary row operation to reduce it. Start by subtracting row 1 from row 3:

\left[ \begin{array}{ccc}<br /> 3 &amp; 1 &amp; 3\\<br /> 1 &amp; 5 &amp; 1\\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}\right]

Next subtract 3 times row two from row 1:

\left[ \begin{array}{ccc}<br /> 0 &amp; -14 &amp; 0\\<br /> 1 &amp; 5 &amp; 1\\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}\right]

This corresponds to the equations
-14 y = 0
x + 5y + z = 0

You get y = 0, z can be anything, say z = c, and x = -z = -c. So you get

\left[ \begin{array}{ccc}<br /> x\\<br /> y\\<br /> z<br /> \end{array}\right] = <br /> <br /> \left[ \begin{array}{ccc}<br /> c\\<br /> 0\\<br /> -c<br /> \end{array}\right] = c\left[ \begin{array}{ccc}<br /> 1\\<br /> 0\\<br /> -1<br /> \end{array}\right]<br />

[Edit] I see vela types faster than I do

 

[cue928]

The three eq's listed as solutions to it are:
x1=c1*e^9t + c2*e^6t + c3*e^2t
x2=c1*e^9t - 2c2*e^6t
x3=c1*e^9t + c2*e^6t -c3*e^2t

With the eigenvector you listed, wouldn't the c1's be different? If I didn't list it earlier, the values in that matrix were for c1.

 

[vela]

I think you've gotten your eigenvalues and matrices mixed up. The eigenvector for the matrix you gave us corresponds to the c₃ terms. Either that or you made another mistake earlier.

 

[cue928]

You use elementary row operation to reduce it. Start by subtracting row 1 from row 3:

\left[ \begin{array}{ccc}<br /> 3 &amp; 1 &amp; 3\\<br /> 1 &amp; 5 &amp; 1\\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}\right]

Next subtract 3 times row two from row 1:

\left[ \begin{array}{ccc}<br /> 0 &amp; -14 &amp; 0\\<br /> 1 &amp; 5 &amp; 1\\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}\right]

This corresponds to the equations
-14 y = 0
x + 5y + z = 0

You get y = 0, z can be anything, say z = c, and x = -z = -c. So you get

\left[ \begin{array}{ccc}<br /> x\\<br /> y\\<br /> z<br /> \end{array}\right] = <br /> <br /> \left[ \begin{array}{ccc}<br /> c\\<br /> 0\\<br /> -c<br /> \end{array}\right] = c\left[ \begin{array}{ccc}<br /> 1\\<br /> 0\\<br /> -1<br /> \end{array}\right]<br />

[Edit] I see vela types faster than I do

So the order of values doesn't matter then?

 

[cue928]

I think you've gotten your eigenvalues and matrices mixed up. The eigenvector for the matrix you gave us corresponds to the c₃ terms. Either that or you made another mistake earlier.

Maybe a dumb question but here goes...I had three values for lambda: 2,6,9. The values in the matrix I listed above were calculated using lambda = 2. Is there a methodology for matching those up?

 

[vela]

The order matters. Note that the eigenvector I wrote down and the one LCKurtz wrote down differ only by a multiplicative constant. Negate mine and you get LCKurtz's. Non-degenerate eigenvectors are unique only up to a multiplicative constant.

 

[cue928]

The order matters. Note that the eigenvector I wrote down and the one LCKurtz wrote down differ only by a multiplicative constant. Negate mine and you get LCKurtz's. Non-degenerate eigenvectors are unique only up to a multiplicative constant.

Ok so how do I set those up? I mean, I see that now, but how could I have discerned that lambda=2 goes with c3 or c2 or c1?

 

[vela]

Maybe a dumb question but here goes...I had three values for lambda: 2,6,9. The values in the matrix I listed above were calculated using lambda = 2. Is there a methodology for matching those up?

Yes, but that's backwards. Normally, you're not going to have the solution to which you match up the values you found.

How do you form a term in the solution once you have an eigenvalue and its associated eigenvector?

 

[cue928]

Yes, but that's backwards. Normally, you're not going to have the solution to which you match up the values you found.

How do you form a term in the solution once you have an eigenvalue and its associated eigenvector?

Wait a sec, I see that now. Okay. I'm still a little uncertain on getting the eigenvectors from the matrices though. You did it one way, the other guy did it with (row reduction?)? What should I do on that? (BTW, I cannot believe LA is not a prereq for ODE's - the terms you threw around earlier I have not heard)

 

[vela]

We both essentially did the same thing. We solved the system of equations you get after substituting the eigenvalue back into (A-λI)x=0. LCKurtz used row reduction. I just told you what the answer was without showing the algebra to solve it. Use whatever method you want to find the solutions to the equations.

 

[cue928]

We both essentially did the same thing. We solved the system of equations you get after substituting the eigenvalue back into (A-λI)x=0. LCKurtz used row reduction. I just told you what the answer was without showing the algebra to solve it. Use whatever method you want to find the solutions to the equations.

On row reduction, what form are you ultimately trying to get it into? Also, does your method always work?

 

[vela]

Into whatever form gets you the solution. The method of solving the system of equations isn't important. You'll get the same answer regardless of the method you choose.

 

[cue928]

That's cute but I saw something that said 1's and 0's in a triangular shape?