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[SOLVED] Jordan Normal Form of a Linear Transformation

Sudharaka

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MHB Math Helper
Feb 5, 2012
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Hi everyone, :)

Here's a question I encountered recently and did partway. I need your advice on how to proceed.

Question:

What can be said about the Jordan normal form of a linear transformation \(f:V\rightarrow V\) where \(V\) is a vector space over \(\mathbb{C}\), if we know that \(f^3=f^2\) ?

My Attempt:

Let \(x\) be a eigenvector of \(f\) and let \(\lambda\) be the corresponding eigenvalue. Than,

\[f(x)=\lambda x\]

\[\Rightarrow f^{2}(x)=\lambda f(x)=\lambda^2 x\Rightarrow f^{3}(x)=\lambda^2 f(x)=\lambda^3 x\]

Since \(f^{3}=f^2\) we have,

\[\lambda^2=\lambda ^3\Rightarrow \lambda =0 \mbox{ or }\lambda =1\]

Now this is where I get stuck. Could you help me out? :)
 

Turgul

Member
Jan 13, 2013
35
We want to understand what Jordan blocks have this behavior. I recommend taking a general 2x2 Jordan form matrix $M$ with eigenvalue $\lambda$ and computing $M^2$ and $M^3$ and seeing what conditions are necessary to make them equal (as you have pointed out, you already know what potentially allowable eigenvalues are, but there is no harm in trying this in general).

Then try it again for a Jordan block of size 3x3. I think you'll get a good picture of what kind of Jordan blocks are allowable (make sure to sort out the 1x1 case too, but that's not too hard).

Since a matrix in Jordan form is simply block diagonal with Jordan block entries, you should be able to get a clean description of all possible Jordan form matrices once you understand what Jordan blocks are allowable.

There exists general machinery to solve this problem quickly, but actually getting your hands a little dirty with this computation should help make the general techniques easier to understand.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
We want to understand what Jordan blocks have this behavior. I recommend taking a general 2x2 Jordan form matrix $M$ with eigenvalue $\lambda$ and computing $M^2$ and $M^3$ and seeing what conditions are necessary to make them equal (as you have pointed out, you already know what potentially allowable eigenvalues are, but there is no harm in trying this in general).

Then try it again for a Jordan block of size 3x3. I think you'll get a good picture of what kind of Jordan blocks are allowable (make sure to sort out the 1x1 case too, but that's not too hard).

Since a matrix in Jordan form is simply block diagonal with Jordan block entries, you should be able to get a clean description of all possible Jordan form matrices once you understand what Jordan blocks are allowable.

There exists general machinery to solve this problem quickly, but actually getting your hands a little dirty with this computation should help make the general techniques easier to understand.
Hi Turgul, :)

Thank you so much for the reply. When I carry on the procedure you have suggested, it seems to me that only \(\lambda =0\) satisfies all the cases. Therefore the Jordan blocks should have \(0\) in their diagonals. Is this true? :)
 

Turgul

Member
Jan 13, 2013
35
Note that, basically due to the computation you did in the original post, the 1x1 blocks must either be 0 or 1, but both work (1x1 Jordan blocks really just being arbitrary complex numbers).

In the n x n case with n > 1, it is necessary that $\lambda = 0$. If not, the super-diagonal entries would differ for $M^2$ and $M^3$. And indeed, in the 2x2 case, it is sufficient. In this case, $M^2$ and $M^3$ should both be the zero matrix. But in the 3x3 case, how do $M^2$ and $M^3$ compare?
 

Sudharaka

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MHB Math Helper
Feb 5, 2012
1,621
Note that, basically due to the computation you did in the original post, the 1x1 blocks must either be 0 or 1, but both work (1x1 Jordan blocks really just being arbitrary complex numbers).

In the n x n case with n > 1, it is necessary that $\lambda = 0$. If not, the super-diagonal entries would differ for $M^2$ and $M^3$. And indeed, in the 2x2 case, it is sufficient. In this case, $M^2$ and $M^3$ should both be the zero matrix. But in the 3x3 case, how do $M^2$ and $M^3$ compare?
How I thought in the 3x3 case was that if we have the Jordan normal form of the matrix as,

\[J=\begin{pmatrix}\lambda&0&0\\0&\lambda&1\\0&0& \lambda\end{pmatrix}\]

then we have,

\[J^2=\begin{pmatrix}\lambda^2&\lambda^2&2\lambda\\0&\lambda^2&2\lambda\\0&0&\lambda^2\end{pmatrix}\]

and

\[J^3=\begin{pmatrix}\lambda^3&\lambda^3&3\lambda\\0&\lambda^3&3\lambda\\0&0&\lambda^3\end{pmatrix}\]

So \(\lambda\) should be equal to zero if \(J^2=J^3\). Is this correct?
 

Turgul

Member
Jan 13, 2013
35
Not quite. Your matrix $J$ should look like:

\[J=\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0& \lambda\end{pmatrix}\]

In general, ALL entries of the super-diagonal should be 1's, so a 4x4 would be:

\[J'=\begin{pmatrix}\lambda&1&0&0\\
0&\lambda&1&0\\
0&0& \lambda&1\\
0&0&0& \lambda \end{pmatrix}\]
 

Sudharaka

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MHB Math Helper
Feb 5, 2012
1,621
Not quite. Your matrix $J$ should look like:

\[J=\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0& \lambda\end{pmatrix}\]

In general, ALL entries of the super-diagonal should be 1's, so a 4x4 would be:

\[J'=\begin{pmatrix}\lambda&1&0&0\\
0&\lambda&1&0\\
0&0& \lambda&1\\
0&0&0& \lambda \end{pmatrix}\]
But isn't that only when we know for sure that there's only one Jordan block for the eigenvalue \(\lambda\)? I mean there could be several Jordan blocks for the same eigenvalue isn't? :)
 

Turgul

Member
Jan 13, 2013
35
And that is exactly why I propose you solve the problem for a Jordan block and then see what Jordan form matrices you can make out of the allowable blocks.

Also, as a caution, your computations of $J^2$ and $J^3$ are not correct for the $J$ you wrote down.
 

Deveno

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MHB Math Scholar
Feb 15, 2012
1,967
We can write a Jordan block as:

$J_{\lambda} = \lambda I +N$

where $N$ is a matrix with all 1's on the super-diagonal and 0's elsewhere.

The matrix $\lambda I$ commutes with any other matrix so:

$(J_{\lambda})^2 = \lambda^2I + 2\lambda N + N^2$

Powers of $N$ just move the 1's "up one diagonal".

We also get:

$(J_{\lambda})^3 = \lambda^3 + 3\lambda^2N + 3\lambda N^2 + N^3$

Comparing the values on the super-diagonal, we get from $(J_{\lambda})^2 = (J_{\lambda})^3$:

$\lambda(3\lambda - 2) = 0$

Since $\lambda \in \{0,1\}$, the root $\lambda = \frac{2}{3}$ (of the equation above) cannot be possible for $n > 1$.
 

Sudharaka

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Feb 5, 2012
1,621
Not quite. Your matrix $J$ should look like:

\[J=\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0& \lambda\end{pmatrix}\]

In general, ALL entries of the super-diagonal should be 1's, so a 4x4 would be:

\[J'=\begin{pmatrix}\lambda&1&0&0\\
0&\lambda&1&0\\
0&0& \lambda&1\\
0&0&0& \lambda \end{pmatrix}\]
And that is exactly why I propose you solve the problem for a Jordan block and then see what Jordan form matrices you can make out of the allowable blocks.

Also, as a caution, your computations of $J^2$ and $J^3$ are not correct for the $J$ you wrote down.
We can write a Jordan block as:

$J_{\lambda} = \lambda I +N$

where $N$ is a matrix with all 1's on the super-diagonal and 0's elsewhere.

The matrix $\lambda I$ commutes with any other matrix so:

$(J_{\lambda})^2 = \lambda^2I + 2\lambda N + N^2$

Powers of $N$ just move the 1's "up one diagonal".

We also get:

$(J_{\lambda})^3 = \lambda^3 + 3\lambda^2N + 3\lambda N^2 + N^3$

Comparing the values on the super-diagonal, we get from $(J_{\lambda})^2 = (J_{\lambda})^3$:

$\lambda(3\lambda - 2) = 0$

Since $\lambda \in \{0,1\}$, the root $\lambda = \frac{2}{3}$ (of the equation above) cannot be possible for $n > 1$.
I am so sorry, I am so tired with this question I made so many mistakes. Now I think I understand what you are trying to say. If we take the 2x2 Jordan blocks and compare \(J^2\) and \(J^3\) we can see it works for both \(\lambda=0\) and \(\lambda=1\). If we take the 3x3 Jordan block,

\[J=\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0& \lambda\end{pmatrix}\]

\[J^3=\begin{pmatrix}\lambda^2&2\lambda&1\\0&\lambda^2&2\lambda\\0&0& \lambda^2\end{pmatrix}\]

and,

\[J=\begin{pmatrix}\lambda^3&3\lambda^2&3\lambda\\0&\lambda^3&3\lambda^2\\0&0& \lambda^3\end{pmatrix}\]

And comparison of \(J^2\) and \(J^3\) would tell us that both \(\lambda=0\) and \(\lambda=1\) are not satisfied in this case. Therefore it seems to me that for Jordan blocks of size 3 and above are non-existant for these \(\lambda\) values. Therefore the possible Jordan blocks are, the size one blocks, \(J_{1}(0)=0,\,J_{1}(1)=1\) and the size two blocks, \(J_{2}(0)=\begin{pmatrix}0&1\\0&0\end{pmatrix},\,J_{2}(1)=\begin{pmatrix}1&1\\0&1\end{pmatrix}\).

Am I correct? :)
 

Klaas van Aarsen

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Sudharaka

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Feb 5, 2012
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Sorry, but no.
Perhaps you can recheck $J_1(1)$?
Well, I don't quite get this. The Jordan block of size 1 which correspond to the eigenvalue \(\lambda=1\) is 1. And of course, $(J_1(1))^2=(J_1(1))^3=1$. So I don't quite understand why this is incorrect? :)
 

Klaas van Aarsen

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Well, I don't quite get this. The Jordan block of size 1 which correspond to the eigenvalue \(\lambda=1\) is 1. And of course, $(J_1(1))^2=(J_1(1))^3=1$. So I don't quite understand why this is incorrect? :)
Sorry, I meant $J_2(1)$.

I tripped and mixed them up.
Note that $J_2$ would usually mean the Jordan block for eigenvalue 2, which is how Deveno used it.
 

Sudharaka

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Feb 5, 2012
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Sorry, I meant $J_2(1)$.

I tripped and mixed them up.
Note that $J_2$ would usually mean the Jordan block for eigenvalue 2, which is how Deveno used it.
I am sorry. I think it's a confusion of notation. You know I wrote $J_{2}(1)$ to mean the Jordan block of size 2 corresponding to the eigenvalue 1. :)
 

Klaas van Aarsen

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I am sorry. I think it's a confusion of notation. You know I wrote $J_{2}(1)$ to mean the Jordan block of size 2 corresponding to the eigenvalue 1. :)
Yes, and I really tried to write using the notation you introduced. (Angel)
 

Sudharaka

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Feb 5, 2012
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Yes, and I really tried to write using the notation you introduced. (Angel)
This notation actually is from Wikipedia (see >>this<<). I fancy it 'cause it gives every detail, the size as well as the corresponding eigenvalue. :)
 

Klaas van Aarsen

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This notation actually is from Wikipedia (see >>this<<). I fancy it 'cause it gives every detail, the size as well as the corresponding eigenvalue. :)
Heh. I surrender. (Tmi)

Anyway, did you recheck the matrix given by a Jordan block of size 2 for eigenvalue 1?
 

Sudharaka

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Feb 5, 2012
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Heh. I surrender. (Tmi)

Anyway, did you recheck the matrix given by a Jordan block of size 2 for eigenvalue 1?
ha ha.... well the Jordan block of size 2 for eigenvalue 1 is,

\[J_{2}(1)=\begin{pmatrix}1&1\\0&1\end{pmatrix}\]

Is there a problem in it? :)
 

Klaas van Aarsen

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ha ha.... well the Jordan block of size 2 for eigenvalue 1 is,

\[J_{2}(1)=\begin{pmatrix}1&1\\0&1\end{pmatrix}\]

Is there a problem in it? :)
Yes. (Headbang) (Wink)
 

Sudharaka

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Feb 5, 2012
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Turgul

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Jan 13, 2013
35
Note that

\[J_{2}(1)^2=\begin{pmatrix}1&2\\0&1\end{pmatrix}\]

and

\[J_{2}(1)^3=\begin{pmatrix}1&3\\0&1\end{pmatrix}\]

Which are not equal. Thus the allowable Jordan blocks are exactly $J_1(0),J_1(1),J_2(0)$ and nothing else. Thus the allowable Jordan form matrices are exactly the block diagonal matrices with blocks any combination of $J_1(0),J_1(1),J_2(0)$.

As Deveno pointed out, powers of $J_m(\lambda)^n$ have a pretty nice general form which I'll write down explicitly for $m=3$

\[J_{3}(\lambda)^n=\begin{pmatrix}
\lambda^n & n \lambda^{n-1} & n(n-1) \lambda^{n-2}\\
0 & \lambda^n & n \lambda^{n-1} \\
0 & 0 & \lambda^n \end{pmatrix}\]

Unless $n=1,2$ where the formula doesn't quite work once 0's end up in the exponents. In fact what actually happens is that, if $f(x) = x^n$, then

\[f(J_3(\lambda)) = J_{3}(\lambda)^n=\begin{pmatrix}
f(\lambda) & f'(\lambda) & f''(\lambda)\\
0 & f(\lambda) & f'(\lambda) \\
0 & 0 & f(\lambda) \end{pmatrix}\]

This patter persists for powering any $J_m(\lambda)$. More generally it is also true for any polynomial $f(x)$.
 

Klaas van Aarsen

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I am sorry, but I don't see the problem. Oh I seem to be so stupid enough not to see this small thing..... (Worried)
We are supposed to have $J^2 = J^3$, but:
$$\begin{pmatrix}1&1\\0&1\end{pmatrix}^2 = \begin{pmatrix}1&2\\0&1\end{pmatrix}$$
$$\begin{pmatrix}1&1\\0&1\end{pmatrix}^3 = \begin{pmatrix}1&3\\0&1\end{pmatrix}$$
 

Deveno

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Feb 15, 2012
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Something well worth remembering (because it turns out to be useful) is that the subgroup:

$G = \left\{\begin{bmatrix}1&k\\0&1 \end{bmatrix}: k \in \Bbb Z\right\}$

of $\text{GL}_2(\Bbb C)$ is isomorphic to the ADDITIVE group of the integers

(or more generally, for $GL_2(V),\ G$ is isomorphic to the additive subgroup of the prime ring of $F$).
 

Sudharaka

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Feb 5, 2012
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Note that

\[J_{2}(1)^2=\begin{pmatrix}1&2\\0&1\end{pmatrix}\]

and

\[J_{2}(1)^3=\begin{pmatrix}1&3\\0&1\end{pmatrix}\]

Which are not equal. Thus the allowable Jordan blocks are exactly $J_1(0),J_1(1),J_2(0)$ and nothing else. Thus the allowable Jordan form matrices are exactly the block diagonal matrices with blocks any combination of $J_1(0),J_1(1),J_2(0)$.

As Deveno pointed out, powers of $J_m(\lambda)^n$ have a pretty nice general form which I'll write down explicitly for $m=3$

\[J_{3}(\lambda)^n=\begin{pmatrix}
\lambda^n & n \lambda^{n-1} & n(n-1) \lambda^{n-2}\\
0 & \lambda^n & n \lambda^{n-1} \\
0 & 0 & \lambda^n \end{pmatrix}\]

Unless $n=1,2$ where the formula doesn't quite work once 0's end up in the exponents. In fact what actually happens is that, if $f(x) = x^n$, then

\[f(J_3(\lambda)) = J_{3}(\lambda)^n=\begin{pmatrix}
f(\lambda) & f'(\lambda) & f''(\lambda)\\
0 & f(\lambda) & f'(\lambda) \\
0 & 0 & f(\lambda) \end{pmatrix}\]

This patter persists for powering any $J_m(\lambda)$. More generally it is also true for any polynomial $f(x)$.
We are supposed to have $J^2 = J^3$, but:
$$\begin{pmatrix}1&1\\0&1\end{pmatrix}^2 = \begin{pmatrix}1&2\\0&1\end{pmatrix}$$
$$\begin{pmatrix}1&1\\0&1\end{pmatrix}^3 = \begin{pmatrix}1&3\\0&1\end{pmatrix}$$
Something well worth remembering (because it turns out to be useful) is that the subgroup:

$G = \left\{\begin{bmatrix}1&k\\0&1 \end{bmatrix}: k \in \Bbb Z\right\}$

of $\text{GL}_2(\Bbb C)$ is isomorphic to the ADDITIVE group of the integers

(or more generally, for $GL_2(V),\ G$ is isomorphic to the additive subgroup of the prime ring of $F$).
Thanks everyone of you for your valuable input on this question. I see my mistakes and see how to approach the problem. :) Thanks so much. :)