When can we say two matrices have the same Jordan form?

Question

When can we say two matrices have the same Jordan form?

Well, it's obvious that if they're similar then they have the same Jordan form, but can we look further?

What if they have the same eigenvalues? That's not enough either because $\begin{bmatrix}0 & 1\\0&0\end{bmatrix}$ and $\begin{bmatrix}0 & 0\\0&0\end{bmatrix}$ have the same values but they are not similar and therefore don't have the same Jordan form.

I can't find an example where two matrices have the same characteristic equation and minimal polynomial yet they are not similar but I am unable to prove it.

Is it enough to have the same algebraic multiplicity and geometric multiplicity?

I am not looking for the minimal condition where two matrices have the same Jordan form

Answer 1

No, there is no weaker condition, and in fact this is the purpose of Jordan normal form: Over an algebraically closed field $\Bbb F$, two matrices are similar iff they have the same Jordan form (up to permutation of Jordan blocks).

Neither of the other conditions mentioned in the question turn out to be sufficient, at least not for matrices of general size.

Let $J_m(\lambda)$ denote the $m \times m$ Jordan block of eigenvalue $\lambda$.

Then, $$J_2(\lambda) \oplus J_2(\lambda) \qquad \textrm{and} \qquad J_2(\lambda) \oplus J_1(\lambda) \oplus J_1(\lambda)$$ both have characteristic polynomial $(t - \lambda)^4$ and minimal polynomial $(t - \lambda)^2$.

Likewise, in both $$J_3(\lambda) \oplus J_1(\lambda) \qquad \textrm{and} \qquad J_2(\lambda) \oplus J_2(\lambda) ,$$ the sole eigenvalue $\lambda$ has geometric multiplicity $4$ and algebraic multiplicity $2$.

These examples are minimal in the sense that both conditions are sufficient for $n \times n$ matrices, $n \leq 3$.

Answer 2

Hint:

If they have the same Jordan form, it means the following:

$$M_1 = {S_1}^{-1}JS_1$$ $$M_2 = {S_2}^{-1}JS_2$$

What can we deduce from that?

Answer 3

Two matrices have the same Jordan form if and only if they are similiar.

Note that the same minimal polynomial and the same characteristic polynomial do not imply that the matrices are similiar.

To investigate this, it suffices to consider nilpotent matrices. Let $n$ be the dimension/size of the matrix.

Then the characteristic polynomial will be $x^n$. Thus the characteristic polynomial will give you no information at all.

Let $d_j = \dim \operatorname{ker} A^j$. We have $0=d_0 < d_1 < d_2 < \dotsb < d_m = n$.

Then the minimal polynomial will be $x^m$. Thus the minimal polynomial will tell you at which point the kernel will be everything.

But from the Jordan form, you can compute all $d_j$. Thus the Jordan form will give you more information then the minimal polynomial and the characteristic polynomial.

For instance you can realize the following sequences with 4x4-matrices:

$0 < 2 < 4$ and $0 < 3<4$

In both cases, the minimal polynomial will be $x^2$, but the first matrix (its Jordan form) will have two blocks of size 2, while the second matrix will have 2 blocks of size 1 and one block of size 2.

Answer 4

Let's suppose for simplicity that two $n$-by-$n$ matrices $A$ and $B$ have only one eigenvalue $\lambda$ (with an additional assumption that the field is algebraically closed). Then, $A$ and $B$ are similar if and only if $\dim\ker\big((A-\lambda I)^k\big)=\dim\ker\big((B-\lambda I)^k\big)$ for all $k=1,2,3,\ldots,n$.

In the case $A$ and $B$ have more than one eigenvalues, you can restrict $A$ and $B$ onto their generalized eigenspaces corresponding to the same eigenvalue $\lambda$. Then, use the previous paragraph to check. However, I don't think that this method is any easier than computing the Jordan normal forms of $A$ and $B$ directly.