Find all the matrices in Jordan form

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SUMMARY

This discussion focuses on finding all matrices in Jordan form with the characteristic polynomial $(x+2)^2(x-5)^3$. The participants identify possible minimal polynomials that divide the characteristic polynomial, including $(x+2)^2(x-5)^3$, $(x+2)(x-5)^3$, and others. They also explore various Jordan forms and their corresponding minimal polynomials, concluding that the Jordan forms can vary based on the arrangement of Jordan blocks. The final forms discussed include matrices with eigenvalues -2 and 5, showcasing different block sizes and arrangements.

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  • Knowledge of eigenvalues and eigenvectors
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  • Learn how to compute minimal polynomials for matrices
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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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