Matrices with given conditions on minimal and characteristic polynomial are similar

Question

Assuming we have two matrices, $A_{1}$ and $A_{2}$. Determine whether they are similar or not given the fact that both matrices satisfies the following conditions (there is no connection between the two sections):

$(1):$ $\chi _{A_{1}}\left( x\right) = \chi _{A_{2}}\left( x\right) = \left( x-1\right) ^{4}x^{2}$ and $\mu _{A_{1}}\left( x\right) = \mu _{A_{2}}\left( x\right) = \left( x-1\right) ^{3}x^{2}$

$(2): dim(ker(\left( A_{1}-I\right) ^{3})) = dim(ker(\left( A_{2}-I\right) ^{3})) = 3$ and $\mu _{A_{1}}\left( x\right) = \mu _{A_{2}}\left( x\right) = \left( x-1\right) ^{3}(x-2)$.

Note that $\mu _{A}\left( x\right)$ is the minimal polynomial of $A$ and $\chi _{A}\left( x\right)$ is the characteristic polynomial of A.

My idea for $(1)$ is that they are similiar because the algebric multiplicity implies that for eigenvalue $\lambda = 1$ the sum of the sizes of the corresponding jordan blocks is 4, and because the multiplicity in the minimal polynomial is 3 so the largest corresponding jordan block is 3, which means that we will have in the jordan matrix $J_{3}\left( 1\right)$ block and $J_{1}\left( 1\right)$ block.

And as for $\lambda = 0$, the largest block will be of size 2 and because the algebric multiplicity is 2 so we will have $J_{2}\left( 0\right)$, and I think that this is the only $6\times 6$ jordan matrix that can satisfies both condintions.

And from here we have that they must be similiar from the uniqueness of jordan form. I think that it is true, but if someone can find a counterexample I'll be glad to know.

and as for $(2)$ I can't find a lead to prove or disprove

Answer 1

Your work and answer for part 1 are correct.

For part 2, the fact that $\dim \ker(A - I)^3 = 3$ that $A$ has on of the following set of blocks associated with the eigenvalue $1$:

• 3 Jordan blocks of size 1,
• 1 Jordan block of size 2 and 1 Jordan block of size 1, or
• 1 Jordan block of size at least 3.

If you want to show that this is the case, I recommend that you show that if $A$ has two blocks with size at least $2$, then $\dim \ker(A - I)^3 > 3$. From there, it is possible to deduce that the outcomes listed above are the only possibilities.

The fact that the exponent of $(x-1)$ in the minimal polynomial of $A$ is $3$ tells you that $A$ has a Jordan block of size 3. Thus, we deduce that $A$ has exactly one Jordan block associated with the eigenvalue $1$ that has size $3$.

The fact that the exponent of $(x-2)$ is $1$ tells you that all Jordan blocks associated with the eigenvalue $2$ have size $1$.

Thus, if it is known a priori that $A_1$ and $A_2$ have the same size, then we can indeed deduce that they must be similar. On the other hand, it is easy to come up with matrices of different sizes that satisfy these conditions.