# Finding the third vector diff equation

1. Aug 11, 2009

### khdani

solve y'=Ay+b system of equation where
$$\vec{A}=\begin{pmatrix} 3 & 1 & 0\\ -1 & 1 & 0\\ 1& 1 &2 \end{pmatrix},\vec{B}=\begin{pmatrix} e^{2x}\\0 \\ e^{2x} \end{pmatrix}$$

i cant see where is my mistake in my method
1st page
http://i26.tinypic.com/24ou8u8.gif
2nd page
http://i29.tinypic.com/1j40ue.gif

2. Aug 11, 2009

### Staff: Mentor

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.

3. Aug 11, 2009

### HallsofIvy

Staff Emeritus
I haven't looked at your attachment but since Mark44 says you have 2 as a triple eigenvalue, the characteristic equation must be $(\lambda- 2)^3= 0$. Now, it is true that a matrix always satisfies its own characteristic equation so, writing "A" for the matrix, we must have $(A- 2I)^3 v= 0$ for all vectors v. Mark44 says you have two independent eigenvectors, say, $v_1$ and $v_2$ so you have $(A- 2I)v_1= 0$ and $(A- 2I)v_2= 0$. From that it follows that $(A- 2I)^2v_1= 0$ and $(A- 2I)^2v_2= 0$. You want to find a third, independent, vector, w, such that neither $(A- 2I)w= 0$ nor $(A- 2I)^2w= 0$ but $(A- 2I)^2w= 0$. Finding a vector, w, such that $(A- 2I)w= v_1$ or $(A- 2I)w= v_2$ will do that.

4. Aug 12, 2009

### khdani

i cant understand how and why you transformed
$$(A- 2I)v_1= 0$$
into
$$(A- 2I)^2v_1= 0$$

and i cant practically understand how to find the third vector from this:"
such that neither $$(A- 2I)w= 0$$ nor $$(A- 2I)^2w= 0$$ but $$(A- 2I)^2w= 0 .$$ Finding a vector, w, such that
$$(A- 2I)w= v_1$$ or $$(A- 2I)w= v_2 will do that.$$"

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.

???