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I want to find all possible Jordan Canonical Forms (JCF) of $3\times 3$ matrix $A$, if $A$ is similar to $A^2$.

I have only $4$, and I think there is something wrong with my solution.

Let $J$ be JCF of $A$, then $J^2$ is JCF of $A^2$, since $A$ and $A^2$ are similar, $J=J^2$. Then the minimal polynomial of $J$ divides $x(x-1)$, so $m(x)$ is one of $x, x-1, x(x-1)$. So: $$J=[0]+[0]+[0], [1]+[1]+[1], [0]+[0]+[1] or [0]+[1]+[1]$$

But, $J$ could also be $[0]+J2[1]$, which is not in the list above. Am I leaving out some cases?

Thanks,

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Please fix your title. Use $\LaTeX$. And what's that J2[1]? –  user21436 Dec 27 '11 at 22:42
    
By J2[1] I meant Jordan block of size 2 corresponding to 1 –  Jennie Dec 27 '11 at 23:08
4  
The square of a matrix in Jordan canonical form is not necessarily again in Jordan canonical form (your $J2[1]$ is an example: compute its square). Therefore your claim "then $J^2$ is the JCF of $A^2$" is incorrect. –  Greg Martin Dec 28 '11 at 0:58
1  
I think another loophole is $[0]+[\omega]+[\omega^2]$, where $\omega$ is a nonreal cube root of unity. –  Gerry Myerson Dec 28 '11 at 2:46
1  
Why wouldn't you allow J3[1] also? Gerry found an interesting subtlety (I think we can use 1 in place of 0 there, also). A further subtlety is that if we are in characteristic two, then [0]+J2[1] and [1]+J2[1] won't work. –  Jyrki Lahtonen Dec 28 '11 at 8:48

1 Answer 1

up vote 3 down vote accepted

The similarity type of a $3\times 3$ matrix is completely determined by its characteristic and minimal polynomials, since they completely determine the Jordan canonical form.

If $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. Since we are assuming that $A$ and $A^2$ are similar, it follows that the multiset of eigenvalues of $A$ (counting multiplicity), $\{\lambda_1,\lambda_2,\lambda_3\}$, equals the multiset $\{\lambda_1^2,\lambda_2^2,\lambda_3^2\}$.

Case 1. $A$ has an eigenvalue that is not $0$ or $1$.

If, say, $\lambda_1\notin\{0,1\}$, then we must have either $\lambda_1^2 = \lambda_2$ and $\lambda_2^2 = \lambda_1$, in which case $\lambda_1^4 = \lambda_1$, so $\lambda$ is a root of $x^4-x=x(x^3-1)=x(x-1)(x^2+x+1)$, hence $\lambda_1$ is a root of $x^2+x+1$, and $\lambda_2$ is the other root; or else $\lambda_1^2=\lambda_2$, $\lambda_2^2 = \lambda_3$, and $\lambda_3^2 = \lambda_1$, in which case $\lambda_1$ is a root of $x^8-x$, hence of $x^6+x^5+x^4+x^3+x^2+x+1$.

In each of these cases, $A$ has three distinct eigenvalues, hence is diagonalizable. So the possibilities are that the Jordan canonical form of $A$ is, up to a reordering of the eigenvalues, $D(0,\omega,\omega^2)$, $D(1,\omega,\omega^2)$, or $D(\zeta,\zeta^2,\zeta^4)$, where $\omega$ is a root of $x^2+x+1$ and $\zeta$ is a root of $x^6+x^5+x^4+x^3+x^2+x+1$. It is easy to verify that each of these matrices is such that $A^2$ is similar to $A$.

Case 2. All eigenvalues of $A$ satisfy $\lambda^2=\lambda$ (i.e., all eigenvalues are either $0$ or $1$).

Case 2.1. $A$ is diagonalizable.

The diagonal cases are $D(0,0,0)$, $D(0,0,1)$, $D(0,1,1)$, and $D(1,1,1)$. Each is similar to its square.

Case 2.2: The Jordan canonical form of $A$ has a single $3\times 3$ block.

If $A$ has a single $3\times 3$ block corresponding to $0$, then the minimal polynomial of $A$ is $t^3$. That is, $A^3=0$; but then $A^4 = 0$, so the minimal polynomial of $A^2$ divides $t^2$, and so the Jordan canonical form of $A^2$ is not a single $3\times 3$ block. So $A$ would not satisfy the requirements of the problem.

If the minimal polynomial of $A$ is $(t-1)^3$, then the minimal polynomial of $A^2$ is either $t-1$, $(t-1)^2$, or $(t-1)^3$. It cannot be $t-1$, since that would imply that $A$ satisfies $t^2-1$, and so the minimal polynomial of $A$ would divide $(t-1)(t+1)$, a contradiction. If the minimal polynomial of $A^2$ is $(t-1)^2$, then $A$ satisfies $(t^2-1)^2 = (t-1)^2(t+1)^2$. If the characteristic of $\mathbf{F}$ is not $2$, this is impossible again, since $(t-1)^3$ does not divide $(t-1)^2(t+1)^2$. So if the characteristic of $\mathbf{F}$ is not $2$, then the minimal polynomial of $A^2$ will be $(t-1)^3$ as well, and so $A$ will be similar to $A^2$. So a single Jordan block associated to $1$ will also be a possible solution.

If the characteristic is $2$, and $$A = \left(\begin{array}{ccc} 1& 1& 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{array}\right),$$ then $$A^2 = \left(\begin{array}{ccc} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right),$$ and it is easy to check that $A^2$ satisfies $(t-1)^2$; thus, the minimal polynomial of $A^2$ is $(t-1)^2$ (it is not $(t-1)$), so $A^2$ is not similar to $A$.

Case 2.3. The Jordan canonical form of $A$ has a $2\times 2$ and a $1\times 1$ Jordan block, both associated to the same eigenvalue.

If the unique eigenvalue of $A$ is $0$, then this means that the minimal polynomial of $A$ is $t^2$, which means that $A^2$ is the zero matrix, which is not similar to $A$.

If the unique eigenvalue of $A$ is $1$, then the minimal polynomial of $A$ is $(t-1)^2$. It is easy to check that if the characteristic of $\mathbf{F}$ is $2$ then $A^2 = I$, so the minimal polynomial of $A^2$ is $t-1$, hence $A^2$ is not similar to $A$; and if the characteristic of $\mathbf{F}$ is not $2$ then $A^2-I\neq 0$ but $(A^2-I)^2 = 0$, so the minimal polynomial of $A^2$ is also $(t-1)^2$, hence $A$ is similar to $A^2$.

Case 2.4. The Jordan form of $A$ consists of a $2\times 2$ block and a $1\times 1$ block, that correspond to distinct eigenvalues.

If the $2\times 2$ block corresponds to $0$, then the minimal polynomial of $A$ is $t^2(t-1)$; it is easy to verify that $A^2=D(0,0,1)$, so the minimal polynomial of $A^2$ is $t(t-1)$, so $A$ is not similar to $A^2$.

If the $2\times 2$ block corresponds to $1$, then the minimal polynomial of $A$ is $t(t-1)^2$. If the characteristic of $\mathbf{F}$ is $2$, then $A^2 = D(1,1,0)$, so the minimal polynomial of $A^2$ is $t(t-1)$ and $A$ is not similar to $A^2$. If the characteristic of $\mathbf{F}$ is not $2$, then the minimal polynomial of $A^2$ is $t(t-1)^2$, so $A$ is similar to $A^2$.

In summary: if $A$ is similar to $A^2$ (over the splitting field of the characteristic polynomial of $A$), then the possible Jordan canonical forms of $A$ are:

  1. Diagonal: up to a permutation of the diagonal entries, $D(0,0,0)$, $D(0,0,1)$, $D(0,1,1)$, $D(1,1,1)$, $D(0,\omega,\omega^2)$, $D(1,\omega,\omega^2)$, or $D(\zeta,\zeta^2,\zeta^4)$, where $D(a,b,c)$ is the diagonal matrix with diagonal entries $a$, $b$, and $c$, $\omega$ is a root of $x^2+x+1$, and $\zeta$ is a root of $x^6+x^5+x^4+x^3+x^2+x+1$.

  2. A $2\times 2$ and a $1\times 1$ Jordan block: none, if the characteristic of the ground field is $2$. If the characteristic of the ground field is not $2$, then the $2\times 2$ Jordan block must be associated to $1$, and the $1\times 1$ block can be associated to either $0$ or $1$.

  3. A single $3\times 3$ Jordan block: none, if the characteristic of the ground field is $2$. If the characteristic of the ground field is not $2$, then a single $3\times 3$ Jordan block associated to $1$.

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