Find all the matrices in Jordan form

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Discussion Overview

The discussion focuses on finding all matrices in Jordan form corresponding to the characteristic polynomial $(x+2)^2(x-5)^3$. Participants explore the possible minimal polynomials and their implications for the Jordan forms, including specific examples and conjectures about the structure of these matrices.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the possible minimal polynomials are those that divide the characteristic polynomial, listing several candidates.
  • Others present specific Jordan forms and question their correctness, suggesting that the minimal polynomial can be inferred from the size of the largest Jordan block.
  • A participant shares examples of matrices and their corresponding characteristic and minimal polynomials, discussing the implications for Jordan forms.
  • Some participants express uncertainty about the derivation of certain Jordan forms and seek clarification on the relationship between minimal polynomials and Jordan block sizes.
  • There are claims about the elementary divisors and their role in determining the Jordan forms, with some participants suggesting corrections to earlier statements regarding these divisors.
  • Participants discuss the ordering of Jordan blocks and how it affects the possible Jordan forms, indicating that multiple arrangements are possible.

Areas of Agreement / Disagreement

Participants generally agree on the need to consider minimal polynomials and their implications for Jordan forms, but multiple competing views remain regarding specific forms and the correctness of certain examples. The discussion remains unresolved on some points, particularly regarding the complete list of Jordan forms and their derivations.

Contextual Notes

Some participants note that the order of Jordan blocks is flexible, which affects the total number of possible Jordan forms. There are also indications that certain mathematical steps or assumptions may be missing or unclear, particularly in the derivation of minimal polynomials from given matrices.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in linear algebra, particularly those studying Jordan forms, characteristic polynomials, and minimal polynomials in the context of matrix theory.

mathmari
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Hey! :o

I want to find all the matrices in Jordan form with characteristic polynomial the $(x+2)^2(x-5)^3$.

Let $\mathcal{X} (x)=(x+2)^2(x-5)^3$.
The possible minimal polynomials $m(x)$ are the ones that $m(x)\mid \mathcal{X} (x)$, so
  1. $(x+2)^2(x-5)^3$
  2. $(x+2)(x-5)^3$
  3. $(x+2)(x-5)^2$
  4. $(x+2)(x-5)$
  5. $(x+2)^2(x-5)^2$
  6. $(x+2)^2(x-5)$
right? (Wondering)

How could we continue to get all the matrices in Jordan form? (Wondering)
 
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I have done the following:

  1. $b(x)=(x+2)^2(x-5)^3=(x+4x+4)(x^3-15x^2+75x-125)=x^5-11x^4+19x^3+115x^2-200x-500$
    $$\begin{bmatrix}
    0 & 0 & 0 & 0 & 500 \\
    1 & 0 & 0 & 0 & 200 \\
    0 & 1 & 0 & 0 & -115 \\
    0 & 0 & 1 & 0 & -19 \\
    0 & 0 & 0 & 1 & 11
    \end{bmatrix}$$
  2. $b_2(x)=(x+2)(x-5)^3=x^4-13x^3+45x^2+25x-250$
    $b_1(x)=(x+2)$
    $$\begin{bmatrix}
    -2 & 0 & 0 & 0 & 0 \\
    1 & 0 & 0 & 0 & 250 \\
    0 & 1 & 0 & 0 & -25 \\
    0 & 0 & 1 & 0 & -45 \\
    0 & 0 & 0 & 1 & 13
    \end{bmatrix}$$
  3. $b_2(x)=(x+2)(x-5)^2=x^3-8x^2+5x+50$
    $b_1(x)=(x+2)(x-5)=x^2-3x-10$
    $$\begin{bmatrix}
    0 & 10 & 0 & 0 & 0 \\
    1 & 3 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & -50 \\
    0 & 0 & 1 & 0 & -5 \\
    0 & 0 & 0 & 1 & 8
    \end{bmatrix}$$
  4. $b_2(x)=(x+2)(x-5)=x^2-3x-10$
    $b_1(x)=(x+2)(x-5)^2=x^3-8x^2+5x+50$
    $$\begin{bmatrix}
    0 & 0 & -50 & 0 & 0 \\
    1 & 0 & -5 & 0 & 0 \\
    0 & 1 & 8 & 0 & 0 \\
    0 & 0 & 0 & 0 & 10 \\
    0 & 0 & 0 & 1 & 3
    \end{bmatrix}$$
  5. $b_2(x)=(x+2)^2(x-5)^2=x^4-6x^3-11x^2+60x+100$
    $b_1(x)=(x-5)$
    $$\begin{bmatrix}
    5 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & -100 \\
    0 & 1 & 0 & 0 & -60 \\
    0 & 0 & 1 & 0 & 11 \\
    0 & 0 & 0 & 1 & 6
    \end{bmatrix}$$
  6. $b_2(x)=(x+2)^2(x-5)=x^3-x^2-16x-20$
    $b_1(x)=(x-5)^2=x^2-10x+25$
    $$\begin{bmatrix}
    0 & 0 & 20 & 0 & 0 \\
    1 & 0 & 16 & 0 & 0 \\
    0 & 1 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & -25 \\
    0 & 0 & 0 & 1 & 10
    \end{bmatrix}$$

Is this correct? (Wondering)
 
mathmari said:
Hey! :o

I want to find all the matrices in Jordan form with characteristic polynomial the $(x+2)^2(x-5)^3$.

Let $\mathcal{X} (x)=(x+2)^2(x-5)^3$.
The possible minimal polynomials $m(x)$ are the ones that $m(x)\mid \mathcal{X} (x)$, so
  1. $(x+2)^2(x-5)^3$
  2. $(x+2)(x-5)^3$
  3. $(x+2)(x-5)^2$
  4. $(x+2)(x-5)$
  5. $(x+2)^2(x-5)^2$
  6. $(x+2)^2(x-5)$
right? (Wondering)

How could we continue to get all the matrices in Jordan form? (Wondering)

Hey mathmari! (Smile)

Isn't a Jordan form for $(x+2)^2(x-5)^3$ for instance:
\begin{bmatrix}
\begin{array}{|cc|}\hline 5 \\ \hline \end{array} && \huge 0\\
&\begin{array}{|cc|}\hline -2 &1 \\ & -2 \\ \hline \end{array} \\
\huge 0&&\begin{array}{|cc|}\hline 5&1\\ & 5 \\ \hline \end{array} \\
\end{bmatrix}
(Wondering)

In this case the minimal polynomial would be $(x+2)^2(x-5)^2$, since the largest Jordan block for $5$ has size $2$. (Thinking)
 
I like Serena said:
Isn't a Jordan form for $(x+2)^2(x-5)^3$ for instance:
\begin{bmatrix}
\begin{array}{|cc|}\hline 5 \\ \hline \end{array} && \huge 0\\
&\begin{array}{|cc|}\hline -2 &1 \\ & -2 \\ \hline \end{array} \\
\huge 0&&\begin{array}{|cc|}\hline 5&1\\ & 5 \\ \hline \end{array} \\
\end{bmatrix}
(Wondering)

How did you find this form? (Wondering)
I like Serena said:
In this case the minimal polynomial would be $(x+2)^2(x-5)^2$, since the largest Jordan block for $5$ has size $2$. (Thinking)

I haven't undestood why this is the minimal polynomial... Could you explain it to me? (Wondering)
 
mathmari said:
How did you find this form? (Wondering)

See wiki about Jordan forms.
It explains it better than I can. (Thinking)
I haven't undestood why this is the minimal polynomial... Could you explain it to me? (Wondering)

The same wiki page explains it.
 
In my notes there is the following example:

$$A=\begin{pmatrix}
2 & -2 & 14 \\
0 & 3 & -7 \\
0 & 0 & 2
\end{pmatrix}$$

$\mathcal{X}_A(x)=|A-xI|=(2-x)^2(3-x)$

$m_A(x)\mid \mathcal{X}_A(x)$

The possible minimal polynomials are
$(x-3)(x-2)$ or $(x-3)(x-2)^2$

To find the minimal one we check if $(A-3I)(A-2I)=0\checkmark$

$m_A(x)=(x-2)(x-3)$

$b_2(x)=m_A(x)=(x-2)(x-3)=x^2-5x+6 \\ b_1(x)=(x-2)$

Therefore $$\begin{bmatrix}
\begin{array}{|cc|}\hline 2 \\ \hline \end{array} && \huge 0\\
\huge 0&&\begin{array}{|cc|}\hline 0&-6\\ 1 & 5 \\ \hline \end{array} \\
\end{bmatrix}$$
If $b=a_0+a_1x+\dots +a_{k-1}x^{k-1}+x^k$ then $\begin{bmatrix}
0 & \dots & 0 & -a_0\\
1 & \ddots & 0 & -a_1\\
0 & \ddots & \ddots & \dots \\
0 & \dots & 1 & -a_{k-1} \\
\end{bmatrix}$
 
mathmari said:
In my notes there is the following example:

$$A=\begin{pmatrix}
2 & -2 & 14 \\
0 & 3 & -7 \\
0 & 0 & 2
\end{pmatrix}$$

$\mathcal{X}_A(x)=|A-xI|=(2-x)^2(3-x)$

$m_A(x)\mid \mathcal{X}_A(x)$

The possible minimal polynomials are
$(x-3)(x-2)$ or $(x-3)(x-2)^2$

To find the minimal one we check if $(A-3I)(A-2I)=0\checkmark$

$m_A(x)=(x-2)(x-3)$

$b_2(x)=m_A(x)=(x-2)(x-3)=x^2-5x+6 \\ b_1(x)=(x-2)$

Therefore $$\begin{bmatrix}
\begin{array}{|cc|}\hline 2 \\ \hline \end{array} && \huge 0\\
\huge 0&&\begin{array}{|cc|}\hline 0&-6\\ 1 & 5 \\ \hline \end{array} \\
\end{bmatrix}$$
If $b=a_0+a_1x+\dots +a_{k-1}x^{k-1}+x^k$ then $\begin{bmatrix}
0 & \dots & 0 & -a_0\\
1 & \ddots & 0 & -a_1\\
0 & \ddots & \ddots & \dots \\
0 & \dots & 1 & -a_{k-1} \\
\end{bmatrix}$

Hey mathmari! (Smile)

It seems a bit complicated to me...
As I see it, the possible Jordan normal forms are:
\begin{bmatrix}
2&&\\
&2&\\
&&3
\end{bmatrix}
\begin{bmatrix}
2&1&\\
&2&\\
&&3
\end{bmatrix}
and beyond that, actually any matrix that has a different ordering of the Jordan blocks.
Btw, they correspond exactly to the minimal polynomials $(x-2)(x-3)$ respectively $(x-2)^2(x-3)$. (Thinking)
 
Oh sorry... I was confused... That what I wrote above is for the rational form of the matrix... (Tmi)

So, to find the Jordan form, we have that the eigenvalues of the original matrix are the following:
$-2, -2, 5, 5, 5$

So, is then the Jordan form the following?
$$\begin{bmatrix}
-2 & 1 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 & 0 \\
0 & 0 & 5 & 1 & 0 \\
0 & 0 & 0 & 5 & 1 \\
0 & 0 & 0 & 0 & 5
\end{bmatrix}$$

(Wondering)
 
That is one of the possible Jordan forms yes. (Nod)
 
  • #10
I like Serena said:
That is one of the possible Jordan forms yes. (Nod)

And how can we find the other ones? (Wondering)

Do we consider the polynomials that I wrote in #1 ? (Wondering)
 
  • #11
mathmari said:
And how can we find the other ones? (Wondering)

Do we consider the polynomials that I wrote in #1 ? (Wondering)

Yes. I gave an example in post #3. (Thinking)
 
  • #12
The elemtary divisors are:
  1. $(x+2)^2(x-5)^3$
  2. $(x+2)(x-5)^3, (x+2)$
  3. $(x+2)(x-5)^2, (x+2)(x-5)$
  4. $(x+2), (x+2), (x-5), (x-5), (x-5)$
  5. $(x+2)^2(x-5)^2, (x-5)$
  6. $(x+2)^2(x-5), (x-5)^2$

Thefore the Jordan forms are the following:
  1. $$\begin{bmatrix}
    -2 & 1 & 0 & 0 & 0 \\
    0 & -2 & 0 & 0 & 0 \\
    0 & 0 & 5 & 1 & 0 \\
    0 & 0 & 0 & 5 & 1 \\
    0 & 0 & 0 & 0 & 5
    \end{bmatrix}$$
  2. $$\begin{bmatrix}
    -2 & 0 & 0 & 0 & 0 \\
    0 & 5 & 1 & 0 & 0 \\
    0 & 0 & 5 & 1 & 0 \\
    0 & 0 & 0 & 5 & 0 \\
    0 & 0 & 0 & 0 & -2
    \end{bmatrix}$$
  3. $$\begin{bmatrix}
    -2 & 0 & 0 & 0 & 0 \\
    0 & 5 & 1 & 0 & 0 \\
    0 & 0 & 5 & 0 & 0 \\
    0 & 0 & 0 & -2 & 0 \\
    0 & 0 & 0 & 0 & 5
    \end{bmatrix}$$
  4. $$\begin{bmatrix}
    -2 & 0 & 0 & 0 & 0 \\
    0 & -2 & 0 & 0 & 0 \\
    0 & 0 & 5 & 0 & 0 \\
    0 & 0 & 0 & 5 & 0 \\
    0 & 0 & 0 & 0 & 5
    \end{bmatrix}$$
  5. $$\begin{bmatrix}
    -2 & 1 & 0 & 0 & 0 \\
    0 & -2 & 0 & 0 & 0 \\
    0 & 0 & 5 & 1 & 0 \\
    0 & 0 & 0 & 5 & 0 \\
    0 & 0 & 0 & 0 & 5
    \end{bmatrix}$$
  6. $$\begin{bmatrix}
    -2 & 1 & 0 & 0 & 0 \\
    0 & -2 & 0 & 0 & 0 \\
    0 & 0 & 5 & 0 & 0 \\
    0 & 0 & 0 & 5 & 1 \\
    0 & 0 & 0 & 0 & 5
    \end{bmatrix}$$

Is this correct? (Wondering)
 
  • #13
Aaalllmost! (Wink)

Item 4 should list $(x+2)(x-5)$ instead of $(x+2)$.
Item 6 should have only size 1 blocks for eigenvalue 5.
Each item should have all possible orderings of the Jordan blocks, since the order of the blocks is left free by the definition. (Nerd)
 
  • #14
I thought about the elementary divisors again... (Thinking)

Are the elementary divisors the following?
  1. $(x+2)^2, (x-5)^3$
  2. $(x+2), (x-5)^3, (x+2)$
  3. $(x+2), (x-5)^2, (x+2), (x-5)$
  4. $(x+2), (x+2), (x-5), (x-5), (x-5)$
  5. $(x+2)^2, (x-5)^2, (x-5)$
  6. $(x+2)^2, (x-5), (x-5), (x-5)$

(Wondering)
 
  • #15
mathmari said:
I thought about the elementary divisors again... (Thinking)

Are the elementary divisors the following?
  1. $(x+2)^2, (x-5)^3$
  2. $(x+2), (x-5)^3, (x+2)$
  3. $(x+2), (x-5)^2, (x+2), (x-5)$
  4. $(x+2), (x+2), (x-5), (x-5), (x-5)$
  5. $(x+2)^2, (x-5)^2, (x-5)$
  6. $(x+2)^2, (x-5), (x-5), (x-5)$

(Wondering)

I'm not entirely sure what you mean by "elementary divisors"... (Thinking)
I guess you mean to identify the sizes of the Jordan blocks.
Then yes, those are the relevant divisors.
 
  • #16
Ah ok... So, the Jordan forms are the ones that I wrote in #12, or not? (Wondering)
 
  • #17
mathmari said:
Ah ok... So, the Jordan forms are the ones that I wrote in #12, or not? (Wondering)

Your matrix in item 6 should have only size 1 blocks for eigenvalue 5.
Currently there is a size 2 block.And I think each item should have all possible orderings of the Jordan blocks, since the order of the blocks is left free by the definition.
So for instance, next to:
$$\begin{bmatrix}
-2 & 0 & 0 & 0 & 0 \\
0 & 5 & 1 & 0 & 0 \\
0 & 0 & 5 & 1 & 0 \\
0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & -2
\end{bmatrix}$$
we should also have:
$$\begin{bmatrix}
-2 & 0 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 & 0 \\
0 & 0 & 5 & 1 & 0 \\
0 & 0 & 0 & 5 & 1 \\
0 &0 & 0 & 0 & 5 \\
\end{bmatrix}$$
(Thinking)
 
  • #18
I like Serena said:
Your matrix in item 6 should have only size 1 blocks for eigenvalue 5.
Currently there is a size 2 block.

Oh yes, it should be:

6. $$\begin{bmatrix}
-2 & 1 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 & 0 \\
0 & 0 & 5 & 0 & 0 \\
0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & 5
\end{bmatrix}$$

right? (Wondering)
I like Serena said:
And I think each item should have all possible orderings of the Jordan blocks, since the order of the blocks is left free by the definition.
So for instance, next to:
$$\begin{bmatrix}
-2 & 0 & 0 & 0 & 0 \\
0 & 5 & 1 & 0 & 0 \\
0 & 0 & 5 & 1 & 0 \\
0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & -2
\end{bmatrix}$$
we should also have:
$$\begin{bmatrix}
-2 & 0 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 & 0 \\
0 & 0 & 5 & 1 & 0 \\
0 & 0 & 0 & 5 & 1 \\
0 &0 & 0 & 0 & 5 \\
\end{bmatrix}$$
(Thinking)

Aren't these two matrices similar? (Wondering)
 
  • #19
mathmari said:
Oh yes, it should be:

6. $$\begin{bmatrix}
-2 & 1 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 & 0 \\
0 & 0 & 5 & 0 & 0 \\
0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & 5
\end{bmatrix}$$

right? (Wondering)

Right. (Nod)

Aren't these two matrices similar? (Wondering)

Yes, so we can leave them out if we're not interested in variants of the same Jordan form that are similar. (Nerd)
 
  • #20
Ah ok... Thanks a lot! (Mmm)
 

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