# Jordan Canonical Form determined by characteristic and minimal polynomials in dimension $3$, but not beyond

Why and how is the Jordan Canonical form of a matrix in $M_3(\mathbb C)$ fully determined by its characteristic and minimal polynomials? And why does it fail for $n >3$?

Thanks.

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If $A\in M_3$ has one eigenvalue $a$, then its characteristic polynomial is $(x-a)^3$, and its minimal polynomial is $x-a$, $(x-a)^2$, or $(x-a)^3$. The only corresponding possibilities for Jordan form are, respectively, 3 blocks of size 1 ($A=aI$), a block of size 2 and hence another of size 1, and one block of size 3. Similarly, the possibilities are determined if there is more than one eigenvalue.

In $M_4$, consider the matrices $\begin{bmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{bmatrix}$ and $\begin{bmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}$.

From the characteristic polynomial $(x-a_1)^{n_1}\cdots (x-a_k)^{n_k}$ of a matrix $A\in M_n$ with distinct eigenvalues $a_1,\ldots,a_k$, you can see that the Jordan blocks for the eigenvalue $a_j$ have to have sizes adding to $n_j$. The largest size of one of these blocks is the exponent of $(x-a_j)$ appearing in the minimal polynomial of $A$. In case $n_j=3$, the only sizes possible for the blocks are $1+1+1$, $2+1$, and $3$, so the highest size of a block (the exponent in the minimal polynomial) determines the decomposition for $a_j$. If $n_j=2$ you have $1+1$ and $2$ as possible Jordan structure, so again it is determined by the largest size of a block. But when $n_j>3$, there are always decompositions like $2+2+[n-4\text{ ones}]$ and $2+1+1+[n-4\text{ ones}]$ which have the same largest size block (same exponent in the minimal polynomial), but are different Jordan forms.

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Thank you very much, Jonas! –  Fred Nov 20 '11 at 0:15

In $M_3(\mathbb{C})$, if your characteristic polynomial gives three distinct complex roots your matrix will be diagonalisable. If you have one complex root and another double root, we can analyse it as follows:

Say your roots of the characteristic equation are $\lambda$ and $\lambda_1$ (twice). Then the block corresponding to the first eigenvalue is already completely determined. Then we only need to concentrate on the second block, which will be $2 \times 2$. This is because if the characteristic polynomial of your matrix is $(t - \lambda)(t - \lambda_1)^2$, there must be the factor $(t - \lambda)$ when $\lambda \neq 0$ in your minimal polynomial, otherwise there is no way to eliminate the block corresponding to $\lambda$ in the Jordan Form of the matrix when $\lambda \neq 0$ (Recall the definition of the minimal polynomial). To see this, observe that if your matrix $M$ is block diagonal, multiplication is done as follows:

$M = \left[\begin{array}{c|c} A & 0 \\ \hline 0 & B \end{array}\right]$, $M^2$ will be equal to $\left[\begin{array}{c|c} A^2 & 0 \\ \hline 0 & B^2 \end{array}\right].$

Having understood this, let us look at the factor in the minimal polynomial corresponding to that block. It will either be $(t - \lambda_1)$ or $(t - \lambda_1)^2$, if $\lambda_1 \neq 0$. Now the first case corresponds to the $2 \times 2$ block being

$\left[\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_1 \end{array}\right]$ (why ?) while the second case to corresponds to a second block of the form $\left[ \begin{array}{cc} \lambda_1 & 1 \\ 0 & \lambda_1 \end{array} \right]$ (why)?

In the case when $\lambda$ or $\lambda_1$ is zero, you should know how the Jordan Form will look like given the minimal and characteristic polynomial just by looking at their definitions.

In the last case that you have three repeated complex roots, $\chi(t)$ will look like $(t - \lambda)^3$. Now we know that the minimal polynomial divides the characteristic polynomial, so that it will either be $(t - \lambda), (t - \lambda)^2$ or $(t - \lambda)^3$. This will then completely the determine the Jordan Form of your matrix, because in the first case, there are no $1's$ on the superdiagonal, the second case only one $1$ on the super diagonal up to rearrangement, and the third case two ones on the super diagonal.

From this we can see that in the case of $3 \times 3$ complex matrices, the characteristic and minimal polynomial of a matrix completely determines its Jordan Form. I believe Jonas has already told you to consider the $4 \times 4$ complex matrices he has given to see why this does not work when $n > 3$.

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The first paragraph is incorrect: the matrix isn't necessarily real so the eigenvalues need not be as described. –  KCd Nov 20 '11 at 0:03
Thanks, everyone! –  Fred Nov 20 '11 at 0:18
@JonasMeyer I have edited my answer. –  user38268 Nov 20 '11 at 0:25
@KCd I was wrong previously. I have now edited my answer. –  user38268 Nov 20 '11 at 0:30
@Fred You're welcome. Do you understand why in each of the three cases above the Jordan Form must look like what I have described? –  user38268 Nov 20 '11 at 0:31