Finding the third vector diff equation

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Homework Help Overview

The discussion revolves around solving a system of differential equations represented by \( y' = Ay + b \), where the matrix \( A \) has repeated eigenvalues. Participants are exploring the implications of having a triple eigenvalue and the associated challenges in finding a complete set of eigenvectors.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for a third independent eigenvector given that the matrix has a triple eigenvalue but only two known eigenvectors. There are inquiries about the transformation of equations and the conditions required for finding the third vector.

Discussion Status

The conversation is ongoing, with some participants providing insights into the nature of the eigenvalues and the requirements for diagonalization. There is a recognition of the need for further exploration to identify the third vector, but no consensus has been reached on the method to achieve this.

Contextual Notes

Participants express confusion regarding the mathematical transformations involved and the implications of the conditions for the third vector. The discussion reflects the complexity of the problem, particularly in the context of linear dependence and the requirements for a basis in the eigenspace.

khdani
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solve y'=Ay+b system of equation where
[tex] \vec{A}=\begin{pmatrix}<br /> 3 & 1 & 0\\<br /> -1 & 1 & 0\\<br /> 1& 1 &2<br /> \end{pmatrix},\vec{B}=\begin{pmatrix}<br /> e^{2x}\\0<br /> \\ e^{2x}<br /> <br /> \end{pmatrix}[/tex]

i can't see where is my mistake in my method
1st page
http://i26.tinypic.com/24ou8u8.gif
2nd page
http://i29.tinypic.com/1j40ue.gif
 
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Your problem is that you have three repeated eigenvalues (lambda = 2), but only two eigenvectors. To be able to diagonalize your matrix (which is what the technique you are using requires), you need to find a basis for the three-dimensional eigenspace, but you have found only two of these basis vectors. Taking a linear combination your two vectors doesn't get you anywhere. No matter what vector you end up with by doing this, your three vectors are linearly dependent, and hence don't form a basis for your eigenspace.

I don't recall what, if anything, you can do in this case, as I'm not anywhere close to my reference books.
 
I haven't looked at your attachment but since Mark44 says you have 2 as a triple eigenvalue, the characteristic equation must be [itex](\lambda- 2)^3= 0[/itex]. Now, it is true that a matrix always satisfies its own characteristic equation so, writing "A" for the matrix, we must have [itex](A- 2I)^3 v= 0[/itex] for all vectors v. Mark44 says you have two independent eigenvectors, say, [itex]v_1[/itex] and [itex]v_2[/itex] so you have [itex](A- 2I)v_1= 0[/itex] and [itex](A- 2I)v_2= 0[/itex]. From that it follows that [itex](A- 2I)^2v_1= 0[/itex] and [itex](A- 2I)^2v_2= 0[/itex]. You want to find a third, independent, vector, w, such that neither [itex](A- 2I)w= 0[/itex] nor [itex](A- 2I)^2w= 0[/itex] but [itex](A- 2I)^2w= 0[/itex]. Finding a vector, w, such that [itex](A- 2I)w= v_1[/itex] or [itex](A- 2I)w= v_2[/itex] will do that.
 
i can't understand how and why you transformed
[tex] (A- 2I)v_1= 0[/tex]
into
[tex] (A- 2I)^2v_1= 0[/tex]

and i can't practically understand how to find the third vector from this:"
such that neither [tex](A- 2I)w= 0[/tex] nor [tex](A- 2I)^2w= 0[/tex] but [tex](A- 2I)^2w= 0 .[/tex] Finding a vector, w, such that
[tex](A- 2I)w= v_1[/tex] or [tex](A- 2I)w= v_2 will do<br /> that.[/tex]"

there is a conflicting conditions in the first sentence.
and i need to do 3 matrix multiplications some with power 2
which is another multiplication.
and i need to see that it differs 0.

?
 

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