Consider an matrix $A_{5x5}$ with only one eigenvalue $\lambda$. If the dimension of the eigenspace $\lambda$ is two, then we can have two possibilities of the Jordan blocs here:

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I didnt understand the statement marked by yellow so I have two questions about it:

1) If $(J^{(2)}-\lambda I)^3$ and $(J^{(3)}-\lambda I)^3$ were equal zero, then $J^{(2)}$ and $J^{(3)}$ would be similar? Why?

2) Like in the reference, why the fact that $(J^{(2)}-\lambda I)^3=0$ and $(J^{(3)}-\lambda I)^3 \ne 0$ guarantees that $J^{(2)}$ and $J^{(3)}$ are not similar?

I dont need a proof, just an understandment.

One can find the cited reference here: http://math.postech.ac.kr/~sungpyo/LinearAlge-2007/Chap8.pdf, and the statement marked by yellow in the bottom of the second page of the reference.

obs: I dont know if the tag "jordan normal form" applies here, since I am studying jordan canonical form (dont really know if there's any difference).

  • $\begingroup$ The Jordan canonical form and Jordan normal form are just different ways of referring to the same thing. Anyway, suppose they were similar. Then $J^{(2)}-\lambda I$ and $J^{(3)}-\lambda I$ would also be similar (why?). Write the equality for the similarity transformation. (In general a similarity transformation looks like $A=CBC^{-1}$.) Raise both sides of it to the 3rd power and simplify. Observe that one side is zero and the other side is not. $\endgroup$ – Ian Aug 23 '16 at 18:49
  • $\begingroup$ For future reference, Jordan Normal Form and Jordan Canonical Form are two different names for the same thing. $\endgroup$ – Christian Aug 23 '16 at 18:50
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    $\begingroup$ @Ian, so you did a proof by contradiction. Got this, if you post this as an answer I will accept/up vote it. $\endgroup$ – gustavoreche Aug 23 '16 at 22:01
  • $\begingroup$ The statement in 1) is false. The statement in 2) (which is used in the yellow statement) is true. $\endgroup$ – Ben Grossmann Aug 24 '16 at 2:48
  • $\begingroup$ In the context here, "canonical" means the Jordan form is unique, while "normal" means that a matrix can always be normalised into the said form. Strictly speaking, Jordan forms are not unique, so I prefer the name "Jordan normal form" to "Jordan canonical form", but the latter is still OK, because Jordan forms are unique up to permutations of Jordan blocks. $\endgroup$ – user1551 Aug 24 '16 at 10:57

For your first question, no. It is still possible for non-similar matrices to both satisfy the condition $(J-\lambda I)^3=0$.

The condition $(J-\lambda I)^3=0$ means that your $J$ has no Jordan block of size 4 or bigger. They matrices could still be non-similar. For example, your $J^{(2)}$ and $J=\lambda I_5$ both have this property but $J^{(2)}$ is not similar to a scalar multiple of the identity.

Maybe this isn't surprising. It is generally easier to find something that doesn't match to show two things are different than to find enough stuff that does match to make sure two things are the same.

For your second question, there are a number of ways to see why these matrices cannot be similar (if the highlighted condition is true for one but not the other). Here's one approach...

What does similar mean? One way to define similarity is that $A$ is similar to $B$ iff both represent the same linear operator just possibly using different bases (i.e. there is some linear operator $T$ such that $[T]_\alpha=A$ and $[T]_\beta=B$ for some bases $\alpha$ and $\beta$ where $[T]_{???}$ denotes a coordinate matrix in basis ???). Changing bases merely conjugates one matrix to another.

Because the coordinate matrix map is linear and respects function composition/matrix multiplication we get: $([T]_{???} -\lambda I)^3 = [(T-\lambda I)^3]_{???}$ so that $(T-\lambda I)^3=0$ iff $([T]_{???}-\lambda I)^3=0$. This means that if one coordinate representation of $T$ satisfies the condition $(x-\lambda)^3=0$, then all coordinate representations must satisfy that condition. So if one matrix satisfies this condition and another doesn't, they cannot be similar.

More generally, we have that the dimension of a nullspace (= kernel) of a linear operator has to match the dimension of the nullspace of any of its coordinate matrices. Let $T$ be a linear operator and $A=[T]_\alpha$ some coordinate matrix for $T$.

Then $\dim(N(A-\lambda I)^k) = \dim(\mathrm{ker}(T-\lambda I)^k)=n_k$. For $k=1$, $N(A-\lambda I)$ and $\mathrm{ker}(T-\lambda I)$ are eigenspaces. So $n_1$ counts how many (linearly independent) eigenvectors $T$ (and $A$) have.

$n_2$ counts how many generalized eigenvectors killed by $(x-\lambda)^2$ we have. So $n_2-n_1$ gives the number of generalized level 2 eigenvectors (killed by $(x-\lambda)^2$ but not by $(x-\lambda)$).

It turns out that each Jordan block (associated with $\lambda$) of size $k$ has one generalized eigenvector of level $\ell$ for each $1 \leq \ell < k$. This means that each Jordan block accounts for exactly one eigenvector (so $n_1$ is the number of Jordan blocks). Then $n_2-n_1$ is exactly the number of Jordan blocks of size $2$ or greater. $n_3-n_2$ is the number of Jordan blocks of size $3$ or greater. etc.

So the numbers $n_1,n_2,\dots$ completely determine the number of Jordan blocks (associated with $\lambda$) and their sizes. Since these $n_k$'s are the same regardless of matrix representation, they are invariants under similarity transform.

This in the end is why your highlighted condition can detect that the matrices are not similar.


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