Suppose $A$ is an 8x8 matrix with entries over $\mathbb{C}$ such that

  1. $\dim \ker(A-2I)=2$
  2. $\dim \ker(A-2I)^2=3$
  3. $\dim \ker(A-3I)=2$
  4. $\dim \ker(A-3I)^2=4$
  5. $\dim \ker(A-3I)^3=5$

Problem - What is the characteristic polynomial of $A$, the minimal polynomial of $A$, the Jordan form and the rational form of $A$?

From (2), (5), the eigenvalues are $2,3$ of multiplicity 3,5. So the minimal polynomial $m(x)$ is $(x-2)^3(x-3)^5$ and must necessarily be the characteristic polynomial? How can I determine if there are 1s on the upper subdiagonal? I would like to see how (1), (3), (4) useful.

  • $\begingroup$ I think from (2) and (5) you get that 2 and 3 are of multiplicities 3 and 5, resepctively, not multiplicities 2 and 3. $\endgroup$ – Gerry Myerson Sep 15 '15 at 9:26
  • $\begingroup$ You're right. This should imply that the characteristic and minimal polynomials are the same. $\endgroup$ – cap Sep 15 '15 at 9:32
  • 1
    $\begingroup$ No. For example, $$B=\pmatrix{2&1&0\cr0&2&0\cr0&0&2\cr}$$ the kernel of $B-2I$ has dimension 2, the kernel of $(B-2I)^2$ has dimension 3, the characteristic polynomial is $(x-2)^3$, the minimal polynomial is $(x-2)^2$. $\endgroup$ – Gerry Myerson Sep 15 '15 at 10:50

From the nested kernels: $$\ker(A-\lambda I)\subset\ker(A-\lambda I)^2\subset\dots\subset\ker(A-\lambda I)^r=\ker(A-\lambda I)^{r+1}=\dots $$ if you set $\;d_i=\dim\ker(A-\lambda I)^i$, then $d_i-d_{i-1}\enspace (i\ge 1$ is equal to the number of Jordan blocks$J_\lambda$ of size $\ge i$.

In particular, $d_1$ is the number of Jordan blocks.

So in the present case, we have $2$ Jordan blocks $J_2$ and $2$ Jordan blocks $J_3$. Furthermore there is

  • $3-2=1$ Jordan block $J_2$ of size $2$,
  • $4-2=2$ Jordan blocks $J_3$ of size $\ge 2$, amongst which $5-4=1$ Jordan block of size $3.

Conclusion: The Jordan canonical for of the matrix is:

enter image description here

From this we see the minimal and characteristic polynomials of $A$ are, respectively:

$$(x-2)^2(x-3)^3, \qquad (x-2)^3(x-3)^5.$$

To compute the Frobenius normal form we need the similarity invariants of the matrix. These are polynomials $P_1, \dots, P_r\enspace(r\le 8)$ such that

  • $P_{i+1}\mid P_i$ for all $i<r$
  • $P_1$ is the minimal polynomial of $A$, $(x-2)^2(x-3)$
  • $P_1\dotsm P_r=\chi_A=(x-2)^3(x-3)^5$

Thus either we have only two similarity invariants: $P_1$ and $P_2=(x-2)(x-3)^2$ or three: $P_1$, $P_2=(x-2)(x-3)$, $P_3=x-3$. Let's set $P_3=1$ in the first case.

We know $P_2P_3$ is the g.c.d. of the minors of $A-xI$ of order $8-2+1=7$ and $P_3$ is the g.c.d. of the minors of order $8-3+1=6$ of the same matrix. The difference between both cases lies in the minors of order $6$ being coprime or not…

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  • $\begingroup$ Thank you. I don't see a nice way to get this into rational form since there may be 1 1x1 block (3) or a 3x3 block corresponding to $(x-2)(x-3)^2$. How can I find it? $\endgroup$ – cap Sep 16 '15 at 0:37
  • $\begingroup$ Which normal form are you speaking of? Frobenius normal form? $\endgroup$ – Bernard Sep 16 '15 at 0:49
  • $\begingroup$ Yes, I can narrow it down to 2 candidates but I am not sure which is correct. $\endgroup$ – cap Sep 16 '15 at 0:51
  • $\begingroup$ @cap: for Frobenius decomposition, I added a criterion based on the minors of the matrix (not very effective by hand, I'm afraid…) $\endgroup$ – Bernard Sep 16 '15 at 2:09

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