# non diagonalisable matrix

• Dec 28th 2012, 03:49 AM
velvet
non diagonalisable matrix
Hello everyone!

Can anyone help me please ? I've got a serious problem in maths with the non-diagonalisable matrices!

A is a matrix
[3 1 0]
[-4 -1 0]
[4 -8 -2]

det |A-λI| = (-2-λ)(λ-1)^2
Thus, we find 2 eigen values:
λ= -2 (order 1)
λ= 1 (order 2)

One eigen vector for λ = -2 is v1 = [0,0,1]

The first eigen vector for λ = 1 is v2= [3, -6, 20]
There is no second independent eigen vector for λ = 1 so the matrix is not diagonalisable!
How can we find A' and P the passage matrix anyway ?

Thanks!! I'm desperate :((((((((((

• Dec 28th 2012, 04:18 AM
Deveno
Re: non diagonalisable matrix
what are you asking here? some matrices aren't diagonalizable...that's just how it is.
• Dec 28th 2012, 04:33 AM
velvet
Re: non diagonalisable matrix
yeah, but despite that our teacher wrote :

Considering e(i) independent vectors that generate the basis, and f(ei) the transformation

f(e(1))=3e(1) - 4e(2) + 4e(3)
f(e(2))=9e(1) - 12e(2) + 12e(3)
=> f(3e(1)) = 9e(1) + (2v(1) -6e(1) - 40e(3)) + 12e(3) = 3e(1) + 2v(1) - 28e(3)
=> f(e(1)) = e(1) + 2/3v1 - 28/3.e(3)

Thus,
A'
[-2 0 -28/3]
[0 1 2/3]
[0 0 1]

P (passage matrix)
[0 3 1]
[0 -6 0]
[1 20 0]

I don't understand what we did there :(
• Dec 28th 2012, 05:48 AM
HallsofIvy
Re: non diagonalisable matrix
The matrix is not diagonalizable but can be put in "Jordan normal form". Since this has eigenvalue -2, of algebraic multiplicity 1, and eigenvalue 1, of algebraic multiplicity 2 but geometric multiplicity 1, the Jordan normal form is
$A'= \begin{bmatrix}-2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$.

To write $A'= P^{-1}AP$, P must have an eigenvector corresponding to eigenvalue -2 as first column, an eigenvector corresponding to eigenvalue 1 as second column, and a generalized eigenvector, corresponding to eigenvalue 1, as third column.
Since eigenvalue 1 has algebraic multiplicity 2, there exist a two dimensional subspace such that $(A- I)^2 v= 0$ for any v in that subspace. Since it has geometric multiplicity 1, there exist only a one dimensional subspace such $(A- I)v= 0$. That means there exist vectors, u, such that $(A- I)^2u= 0$ but $(A- I)u\ne 0$. Those are the "generalized eigenvectors" corresponding to eigenvalue 1. We can, however, write $(A- I)^2u= (A- I)[(A- I)u]= (A- I)v= 0$ where $v= (A- I)u$. And since (A- I)v= 0, v must be an eigenvector.

You have already determined that v= <3, -6, 20> (I did not check if this is correct) is an eigenvector corresponding to eigenvalue 1 so any "generalized eigenvector", <x, y, z>, must satisfy
$\begin{bmatrix} 2 & 1 & 0 \\ -4 & -2 & 0 \\ 4 & -8 & -3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3 \\ -6 \\ 20\end{bmatrix}$

Solve that for <x, y, z> (there is not, of course, a unique solution- choose any one) and use it as the third column in your matrix P.
• Dec 28th 2012, 05:49 AM
ILikeSerena
Re: non diagonalisable matrix
Hi velvet! :)

I don't understand either what your teacher did here.
What is e(2) supposed to be?

However, I can tell you how to find an alternate A' and P.
It is called the Jordan normal form and its Generalized eigenvectors (link included).
I'm pretty sure that what you're learning leads up to the Jordan normal form.

See the link I just mentioned for the method how to construct the passage matrix.

Thus,
A' (jordan normal form)
[-2 0 0]
[0 1 1]
[0 0 1]

P (passage matrix containing generalized eigenvectors)
[0 3 11/5]
[0 -6 -7/5]
[1 20 0]
• Dec 28th 2012, 08:42 AM
velvet
Re: non diagonalisable matrix
Ok, I think that the method with the Jordan normal form is quite similar to what we have done.
Thank you all for your answers, I'm starting to get the thing now !
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