Diagonalisable or not ?(Error in question ?)

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http://dl.dropbox.com/u/33103477/pap.png [Broken]

I've worked out the first two bit's which are easy enough:


i)Multiply and get eigenvalues
ii)Turns out that the eigenvalue = 2 has multiplicity 2(algebraic !?!) I did this by row reduction at each eigenvalue to check dimension of solution set.
iii) Now this bit seems impossible as there do not seem to be enough eigenvector's. I've plugged the matrix into a few programs and they can't quite solve it.

So does anyone know what's going on ?

Here is some data so that you don't have to work out anything:

Characteristic polynomial:
[tex] x^4 - 3x^3 - 2x^2 + 12x - 8 [/tex]

Real eigenvalues:
{-2, 1, 2, 2}

Eigenvector of eigenvalue -2:
(1, 1, 0, 0)
Eigenvector of eigenvalue 1:
(0, 0, -1, 1)
Eigenvector of eigenvalue 2:
(1, 0, -1, 1)
Eigenvector of eigenvalue 2:
(1, 0, -1, 1)


The best I can do is say that diagonal of T is made up of the eigenvalues of A.
 
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Answers and Replies

  • #2
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I haven't tried it but it depends whether the eigenvalue two has two linearly independent eigenvalues or not- getting a computer to find one is not going to help here. You need to write out the set of simultaneous equations

A(v) = 2v

and see if when solving them you get two linearly independent possible vectors for v.
 
  • #3
lanedance
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Its worth chekcing whether the matrix has a full set of eigenvectors as suggested.

However, note the question asks for triangularisation, not diagonalisation. If there were a full set of eigenvectors, then we would be able to diagonalise A.

Now in the digonalisable case, we use the eignevctors as the column vectors of P. Note that as the eigenvectors form a basis, P is invertible.

Note that in this case as we require an invertible P, the column vectors of P must fulfill a similar requirement. Any thoughts on what we could attempt to use?
 
  • #5
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Yes, but if you read through the conversation it had gone completely off track.(It became a discussion about types of multiplicities)
When I wanted to work on that problem today I could not locate that old thread so I started a new one.
 
  • #6
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Its worth chekcing whether the matrix has a full set of eigenvectors as suggested.

However, note the question asks for triangularisation, not diagonalisation. If there were a full set of eigenvectors, then we would be able to diagonalise A.

Now in the digonalisable case, we use the eignevctors as the column vectors of P. Note that as the eigenvectors form a basis, P is invertible.

Note that in this case as we require an invertible P, the column vectors of P must fulfill a similar requirement. Any thoughts on what we could attempt to use?

Yes the fact that P has to be invertible has been the biggest stumbling block. I have really no idea how to construct one with only 3 independant eigenvectors.
 
  • #7
lanedance
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Well you have 3 eigenvectors that will be linearly independent as they are from 3 different eigenvalues. As discussed previously you should confirm that the eigenvalue 2 has geometric multicplicity of 1.

Now to make a guess at an invertible matrix P, you require 4 linearly independent vectors.

Assuming you plan to use the eigenvectors as 3 of them, any idea on what to use as the forth?

I should add, I haven't tried the calc yet, but am just trying to add some ideas of ways to approach it.
 
  • #8
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Okay, what I decided to do was

[tex] (A-2I)x=(1,0,-1,1)^T [/tex]

and similarly for all the other eigenvalues. Interestingly there were no solutions for any of the other eigenvalues except for the one with the equation given above.

I got

[tex]

\begin{bmatrix}
4 & -4 & -3 & -3\\
0 & 0 & -4 & -4 \\
-1 & 1 & 4 & 1 \\
1 & -1 & 0 & 3
\end{bmatrix}

\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix} = \begin{bmatrix}
1\\
0\\
-1\\
1
\end{bmatrix}
[/tex]

Which solved to give the following solution:

[tex] x_2=-1, x_3=1-x_1, x_1=x_4 [/tex]

So does this make sense ?
 
  • #9
lanedance
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can you explain what you're attempting?

if you're solving for the eigenvectors relating to the eigenvalue 2, shouldn't it go
Ax=2x=2Ix

giving
Ax-2Ix=0
(A-2I)x=0
 
  • #10
lanedance
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ok, so I just read the post Mark gave. I suggest you first attempt finding the characteristic equation, then solving my last post. This should tell you both the algebraic and geometric multiplicities are. From there we should be in clear place to come up with a method of diagonalisation, or if that's not possible, triangularisation.

Do you understand what we mean by algebraic and geometric multiplicity?
 
  • #12
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can you explain what you're attempting?

if you're solving for the eigenvectors relating to the eigenvalue 2, shouldn't it go
Ax=2x=2Ix

giving
Ax-2Ix=0
(A-2I)x=0

My understanding is algebraic multiplicity is the power of the factors in the characteristic equation.
The geometric multiplicity is a bit more confusing. Basically I think of it as the representation of the vector in a geometric plane. Thus a single eigenvalue could have multiple corresponding eigenvectors and the number of the eigenvectors is the geometric multiplicity.

Now I did what you said (A-2I)x=0 and got :

x1=x2, x3=0, x4=0

Does this mean that any eigenvector satisfying the above can form a geometric multiplicity ? And if so can I use this to form the last eigenvector with any value or a particular value.
 
  • #13
lanedance
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My understanding is algebraic multiplicity is the power of the factors in the characteristic equation.
The geometric multiplicity is a bit more confusing. Basically I think of it as the representation of the vector in a geometric plane. Thus a single eigenvalue could have multiple corresponding eigenvectors and the number of the eigenvectors is the geometric multiplicity.
The algebraic multiplicity is the power of the factored eigenvalue in the characteristic equation. The sum of all the eigenvalue algebraic multiplicities is equal to the dimension of the space, in this case 4.

The geometric multiplicity is the dimension of the associated eigenspace, ie the number of linearly independent eigenvectors in a basis for the eigenspace corresponding to a given eigenvalue. As you mention when dealing with normal vectors, the eigen space can be thought of a a line (dim=1), plane(dim=2) etc. The geometric multiplicity is always less than or equal to the alegebraic multiplicity.

If a matrix has geometric multiplicity equal to algebraic multiplicity for all eigenvalues, there a basis of eigenvectors that spans the whole space, and the matrix is diagonalisable

When a matrix has a geometric multiplicity less than an algebraic multiplicity, there is no set of eigenvectors than spans the whole space (in this case R^4) and the matrix is said to be defective and is not diagaonalisable.
 
  • #14
lanedance
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Now I did what you said (A-2I)x=0 and got :

x1=x2, x3=0, x4=0

Does this mean that any eigenvector satisfying the above can form a geometric multiplicity ? And if so can I use this to form the last eigenvector with any value or a particular value.

If you solved this correctly you should get the eigenvector(s) corresponding to the eigenvalue of 2.

The eiegnevector you found is c(1,1,0,0), for some constant c. The result looks like the eigenvector corresponding to -2, based on your intial post. This would come form solving (A+2I)x=0 instead.

As a test try multiplying the matrix A by the eigenvalue you found (1,1,0,0) and see how it behaves
 
  • #15
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If you solved this correctly you should get the eigenvector(s) corresponding to the eigenvalue of 2.

The eiegnevector you found is c(1,1,0,0), for some constant c. The result looks like the eigenvector corresponding to -2, based on your intial post. This would come form solving (A+2I)x=0 instead.

As a test try multiplying the matrix A by the eigenvalue you found (1,1,0,0) and see how it behaves

You are right I calculated based on the wrong eigenvalue. Doing Linear Algebra and Real Analysis at the same time is doing my head in !

Anyway I recalculated it properly
and I've got x1=x4, x2=0, x3=-x1 which is the matrix of the form:

[tex] c\begin{bmatrix}
1\\
0\\
-1\\
1
\end{bmatrix}[/tex]

Which is the same as given in the question.

So I've come in one enormous circle it seems.
 
  • #17
lanedance
Homework Helper
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You are right I calculated based on the wrong eigenvalue. Doing Linear Algebra and Real Analysis at the same time is doing my head in !

Anyway I recalculated it properly
and I've got x1=x4, x2=0, x3=-x1 which is the matrix of the form:

[tex] c\begin{bmatrix}
1\\
0\\
-1\\
1
\end{bmatrix}[/tex]

Which is the same as given in the question.

So I've come in one enormous circle it seems.

Yes and no. Hopefully you understand multiplicities and solving for eigenvectors now. You could have guessed form the question the matrix was defective and could not be diagonalised, but you've proved it.

So, you have proven to yourself the eigenvalue 2 has only one eigenvector. So the eigenvalue 2 has geometric multiplicity =1.

Now as you know the chracteristic equation is
[tex]
x^4 −3x^3 −2x^2 +12x−8 = (x+2)(x-1)(x-2)^2
[/tex]

The eigenvalue 2 has algebraic multiplicity =2. Hence the matrix is defective and cannot be diagonalised.

We can however find the Jordan Normal form, which is how to find P to transform the matrix into upper triangular form as requested in the question. see Mark's link in the last post
 
  • #18
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Thank you all !! That Wiki page was what I needed managed to work it out perfectly !
 

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