Finding the third vector diff equation

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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.

?