Find all the matrices in Jordan form

In summary, the conversation discusses finding all possible matrices in Jordan form with a specific characteristic polynomial. The process involves finding the minimal polynomial, which is then used to construct the Jordan blocks in the form of a matrix. The possible Jordan forms are determined by the different orderings of the Jordan blocks, which correspond to different minimal polynomials.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

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)
 
Physics news on Phys.org
  • #2
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)
 
  • #3
mathmari said:
Hey! :eek:

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)
 
  • #4
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)
 
  • #5
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.
 
  • #6
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 $\begin{bmatrix}
0 & \dots & 0 & -a_0\\
1 & \ddots & 0 & -a_1\\
0 & \ddots & \ddots & \dots \\
0 & \dots & 1 & -a_{k-1} \\
\end{bmatrix}$
 
  • #7
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)
 
  • #8
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)
 
  • #9
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)
 

Similar threads

I Congruence for Symmetric and non-Symmetric Matrices for Quadratic Form
Replies
7
Views
1K
MHB -m99.54 List possible Jordan Canonical forms for A.
Replies
11
Views
2K
MHB What Are the Possible Jordan Forms and Characteristic Polynomial of T?
Replies
1
Views
2K
MHB Matrices.......whose null space consists all linear combinations
Replies
3
Views
2K
MHB What Are the Minimal Polynomials of Matrix Powers?
Replies
4
Views
2K
I How can I find all possible Jordan forms?
Replies
8
Views
2K
MHB Calculating Determinants: Using Laplace Expansion or Echelon Form?
Replies
13
Views
2K
MHB 25.1 find all possible Jordan Normal Forms of A
Replies
1
Views
2K
MHB Change the form of equation of surface
Replies
6
Views
1K
MHB Set is closed as for multiplication of matrices
Replies
16
Views
2K

Hot Threads

Recent Insights

Back
Top