# Jordan Form. $A\in M_{7}(\mathbb{C})$ and $\operatorname{rank}(A-\lambda I)^2=1$ and $\operatorname{rank}(A-\lambda I)=2$

Let $A\in M_{7}(\mathbb{C})$. $A$ has single eigenvalue $\lambda \in \mathbb{C}$.

Let $\operatorname{rank}(A-\lambda I)^2=1$ and $\operatorname{rank}(A-\lambda I)=2$ .

I need to find $M_{A}$ (The minimal polynomial) and $J_{A}$ (Jordan form for the matrix $A$).

First, I know that with $\ker (A-\lambda I)$ I can find the eigenspace that belongs to the eigenvalue $\lambda$, but what does the rank means in this case? Does it mean the dimension of the image while $\dim$ im $A$ + $\dim \ker$A$= 7$?

What does $\operatorname{rank}(A-\lambda I)^k$ mean? what does $k$ stand for? What is the relation between $\operatorname{rank}(A-\lambda I)^k$ and $\operatorname{rank}(A-\lambda I)^{k-1}$?

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I assume you mean what is the relation between $rank(A-\lambda I)^k$ and $rank(A-\lambda I)^{k-1}$, right? – Manos Jul 25 '11 at 12:13
Yeah, I fixed it. Thanks! – user6163 Jul 25 '11 at 12:20
@Dylan: But there is a clear relationship between the rank of $B$ and $B^2$ if $B$ has $0$ as unique eigenvalue, as i explain in my answer. – Manos Jul 25 '11 at 16:25
@Dylan: I understand that, but does it have anything to do with the minimal polynomial? Does it tell me something about the amount of jordan blocks and etc? this is what I meant. – user6163 Jul 25 '11 at 17:00
@Nir: When you raise $A - \lambda I$ to the $k^{th}$ power, then every nontrivial Jordan block that corresponds to a zero eigenvalue loses $k$ from its rank. The sequence $\rho_k=rank[(A-\lambda I)^k]$ is then used to determine the number and size of Jordan blocks corresponding to eigenvalue $\lambda$. – Manos Jul 25 '11 at 18:01

First observe that the matrices $A$ and $A- \lambda I$ have the same Jordan structure, except from the fact that $A$ has $\lambda$ as its unique eigenvalue with algebraic multiplicity $7$ and $A- \lambda I$ has zero as its unique eigenvalue with algebraic multiplicity $7$. Now, assume that $\lambda \neq 0$ and so $rank(A)=7$. By subtracting $\lambda I$ from $A$, the rank of the resulting matrix $A-\lambda I$ will be as many times less than 7 as the number of Jordan blocks that it has. For example if $A$ is diagonalizable, then its Jordan blocks will be trivial, i.e. each one equal to the $1 \times 1$ matrix $\lambda$ and so $rank(A-\lambda I)=0$. The fact that $rank(A-\lambda I)=2$ tells you that you have at most two nontrivial Jordan blocks, i.e. of size at least $2 \times 2$. Another fact that we need at this point, and it can be checked easily, is that if $J_n(0)$ is a Jordan block of size $n \times n$ with eigenvalue $0$, then $rank(J_n(0)^k)=rank(J_n(0)^{k-1})-1$. This implies that the rank of $(A-\lambda I)^2$ will be as many times less than the rank of $A-\lambda I$ as the number of nontrivial Jordan blocks of $A-\lambda I$ (which are the same number as the nontrivial Jordan blocks of $A$). The fact that $rank(A-\lambda I)-rank((A-\lambda I)^2)=1$, implies that $A$ has only one nontrivial Jordan block of size at least $2 \times 2$. We easily then see that this block must be a $3 \times 3$ and so the minimal polynomial is $m_A(x)=(x-\lambda)^3$ and that $A$ is similar to the matrix $\Lambda= diag(J_1(\lambda), J_1(\lambda), J_1(\lambda), J_1(\lambda), J_3(\lambda))$.

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Here is a simple recipe.

Let $\mathrm{rk}(C)$ denote the rank of $C$, and put $B:=A-\lambda I$. Write the sequence $\mathrm{rk}(B^0)$, $\mathrm{rk}(B^1)$, $\mathrm{rk}(B^2)$, ... [Remember: $B^0=I$, $B^1=B$.] You get $$7,\quad 2, \quad 1,\quad 0,\quad 0,\quad\cdots$$ Take the successive differences: $7-2=5$, $1-0=1$, and so on: $$5,\quad 1, \quad 1,\quad 0,\quad 0,\quad\cdots$$ Do it again: $$4,\quad 0, \quad 1,\quad 0,\quad 0,\quad\cdots$$ This tells you that you have $4$ Jordan blocks of size $1$, $0$ Jordan blocks of size $2$, and $1$ Jordan block of size $3$, and that's it.

More generally, if you put $f(k):=\mathrm{rk}(B^k)$ and $(\Delta g)(k):=g(k)-g(k+1)$, then the number of Jordan blocks of size $k$ is $(\Delta^2 f)(k)$.

Here is a justification. [I'll use the letter $i$ as a subscript; this is not $\sqrt{-1}$.]

Let $A$ be an $n$ by $n$ complex nilpotent matrix, put $V:=\mathbb C^n$, and view $V$ as a $\mathbb C[X]$-module via the formula $fv:=f(A)v$ for $f\in\mathbb C[X]$ and $v\in\mathbb C^n$. Consider the $\mathbb C[X]$-module $$V_i:=\mathbb C[X]/(X^{i+1}).$$ [For any $f\in\mathbb C[X]$, $(f)$ denotes the ideal generated by $f$.] The endomorphism $v\mapsto Xv$ of $V_i$ has just one Jordan block $J(i+1,0)$ of size $i+1$.

By the Jordan block theory, we have $$V\simeq\bigoplus_jV_{i(j)},$$ where the sum is finite. Write this symbolically as $$V\simeq\bigoplus_i\ m_i\ V_i,$$ where $m_i$ is the number of times $V_i$ appears in the previous sum, that is, the number of Jordan blocks of size $i+1$. Set $$n_j:=\dim\frac{X^jV}{X^{j+1}V}\quad.$$

It suffices to verify the

CLAIM: $m_i=n_i-n_{i+1}$.

To prove the claim, form the polynomial $$V(Y):=\sum\ n_j\ Y^j.$$ We have $$V_i(Y)=\frac{Y^{i+1}-1}{Y-1}=1+Y+Y^2+\cdots+Y^i,$$ and we must solve $$\sum\ m_i\ V_i(Y)=\sum\ n_j\ Y^j$$ for the $m_i$, where the $n_j$ are considered as known quantities (almost all equal to zero). Multiplying through by $Y-1$ we get $$\sum\ m_{i-1}\ Y^i-\sum\ m_i=\sum\ (n_{i-1}-n_i)\ Y^i.$$ This proves the claim.

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