jordan forms

**Juneu436** · October 1st 2011, 06:44 AM

Hey, I am having some trouble doing these jordan form questions.

1. What are all the possible Jordan forms for a matrix whose characteristic polynomial is $(\lambda+2)^2(\lambda-5)^3$ .

2. same as Q1 but the space of eigenvectors with eigenvalue 2 is 1-dimensional and is 2-dimensional for eigenvalue 5.

Thanks for any help.

**Juneu436** · October 2nd 2011, 06:19 AM

So for Q1 I know the roots are -2,-2,5,5,5 so they will form the diagonal of the jordan matrix. But how do I find the jordan block/s?

**TheEmptySet** · October 2nd 2011, 07:34 AM

Originally Posted by Juneu436

Hey, I am having some trouble doing these jordan form questions.

1. What are all the possible Jordan forms for a matrix whose characteristic polynomial is $(\lambda+2)^2(\lambda-5)^3$ .

2. same as Q1 but the space of eigenvectors with eigenvalue 2 is 1-dimensional and is 2-dimensional for eigenvalue 5.

Thanks for any help.

You need to make the connection with the minimum polynomial. The minimum polynomial must divide the characteristic polynomial so it must have the form

$m(\lambda)=(\lambda+2)^j(\lambda-5)^k$

Where $j=1, \text{ or }2$ and $k=3,4, \text{ or } 5$

The 2nd fact you need is that the degree of the factor in the minimum polynomial tells you the size of the largest Jordan block associated with that eigenvalue.

So for example if j=1. Then you will have two 1 by 1 blocks for the eigenvalue -2, but if j=2 you would only get one 2 by 2 jordan block. Since you need to find all possible forms you need to let j and k take on all possible values. I hope this clears things up a bit.

**Juneu436** · October 2nd 2011, 04:53 PM

Originally Posted by TheEmptySet

You need to make the connection with the minimum polynomial. The minimum polynomial must divide the characteristic polynomial so it must have the form

$m(\lambda)=(\lambda+2)^j(\lambda-5)^k$

Where $j=1, \text{ or }2$ and $k=3,4, \text{ or } 5$

Thanks, but why is $k=3,4, \text{ or } 5$ ?
Shouldn't it be $k=1,2, \text{ or } 3$ ?

So for question 1 I get:
$m(\lambda)=(\lambda+2)(\lambda-5)$
$\mathbf{J_1} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)^2(\lambda-5)$
$\mathbf{J_2} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)(\lambda-5)^2$
$\mathbf{J_3} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)^2(\lambda-5)^2$
$\mathbf{J_4} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)^2(\lambda-5)^3$
$\mathbf{J_5} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)(\lambda-5)^3$
$\mathbf{J_6} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

Is this correct?

What about question 2?

Thanks

**Juneu436** · October 3rd 2011, 07:50 AM

How should I view 'eigenvectors with eigenvalue 2 is 1-dimensional and is 2-dimensional for eigenvalue 5'? Do I

**TheEmptySet** · October 3rd 2011, 09:15 AM

Originally Posted by Juneu436

Thanks, but why is $k=3,4, \text{ or } 5$ ?
Shouldn't it be $k=1,2, \text{ or } 3$ ?

So for question 1 I get:
$m(\lambda)=(\lambda+2)(\lambda-5)$
$\mathbf{J_1} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)^2(\lambda-5)$
$\mathbf{J_2} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)(\lambda-5)^2$
$\mathbf{J_3} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)^2(\lambda-5)^2$
$\mathbf{J_4} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)^2(\lambda-5)^3$
$\mathbf{J_5} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$m(\lambda)=(\lambda+2)(\lambda-5)^3$
$\mathbf{J_6} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

Is this correct?

What about question 2?

Thanks

Yes you are correct k should be 1,2 or 3.

Question two is just telling you how many linearly independant eigenvectors you get for each space.

Since you get only one linearly independant eigenvector for the eigenvalue -2 that means you will need one generalized eigenvector. This will give a Jordan block of the form.

$\begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}$

The 2nd eigenvalue admits two linearly independant eigenvectors so you will need one generalized eigenvector to finish the form. So what would your blocks look like?

**Juneu436** · October 3rd 2011, 12:16 PM

Originally Posted by TheEmptySet

Yes you are correct k should be 1,2 or 3.

Question two is just telling you how many linearly independant eigenvectors you get for each space.

Since you get only one linearly independant eigenvector for the eigenvalue -2 that means you will need one generalized eigenvector. This will give a Jordan block of the form.

$\begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}$

The 2nd eigenvalue admits two linearly independant eigenvectors so you will need one generalized eigenvector to finish the form. So what would your blocks look like?

Thanks, so this is the only possible jordan form for question 2?
$\mathbf{J} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

**TheEmptySet** · October 3rd 2011, 12:56 PM

Originally Posted by Juneu436

Thanks, so this is the only possible jordan form for question 2?
$\mathbf{J} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

Be careful. When you look at the block for 5 you have

$\begin{bmatrix}5 & 1& 0 \\ 0 & 5 & 1 \\ 0 & 5 & 5 \end{bmatrix}$ This is a 3 by 3 block and so it would only have 1 eigenvector. Remember each Jordan block has exactly one eigenvector in it. So you would have one 2 by 2 block and a 1by 1 block

**Juneu436** · October 3rd 2011, 01:10 PM

Oh, how stupid of me.
So for q2 the only jordan form is
$\mathbf{J} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 5 & 5\end{bmatrix}$

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