# Finding minimal, characteristic polynomials and rational, jordan forms from the dimensions of nullspaces

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.

• I think from (2) and (5) you get that 2 and 3 are of multiplicities 3 and 5, resepctively, not multiplicities 2 and 3. – Gerry Myerson Sep 15 '15 at 9:26
• You're right. This should imply that the characteristic and minimal polynomials are the same. – cap Sep 15 '15 at 9:32
• 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 $(x-2)^2$. – 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 $3. Conclusion: The Jordan canonical for of the matrix is: From this we see the minimal and characteristic polynomials of$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… • 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? – cap Sep 16 '15 at 0:37