# Jordan form of a power of Jordan block?

How, in general, does one find the Jordan form of a power of a Jordan block?

Because of the comments on this question I think there is a simple answer.

• Hint: consider the eigenvalues of the powers of the Jordan block. – amd Jul 5 '16 at 16:03

## 1 Answer

Let $J$ be the $n\times n$ Jordan block with eigenvalue $\lambda$. I'll assume we're working over $\mathbb C$ (or at least in characteristic $0$).

Claim: If $\lambda\neq 0$ then the Jordan normal form of $J^m$ is an $n\times n$ Jordan block with eigenvalue $\lambda^m$. If $\lambda=0$ then the Jordan normal form of $J^m$ is $r$ blocks of size $q+1$ and $m-r$ blocks of size $q$, where $m$ divides $q$ times into $n$ with remainder $r$.

Proof:

Write $J=\lambda I +N$ where $N$ contains ones on the first off-diagonal. Note that $N^m$ is the matrix with ones on the $m$th diagonal away from the main diagonal. So $N^m\neq 0$ for $m<n$ but $N^n=0$.

First consider the case $\lambda\neq 0$. By binomial expansion we have

$$J^m=\sum_{i=0}^m \binom mi \lambda^{m-i}N^{i}.$$

So $J^m$ has $\lambda^m$ along the main diagonal, $m\lambda^{m-1}$ along the next diagonal, and so on. In particular, $J^m$ is triangular with $\lambda^m$ on the diagonal, so each of its eigenvalues is $\lambda^m$.

To discover the size of the largest Jordan block for $J^m$ we should look at which $k$ makes $(J^m-\lambda^m I)^k$ vanish (because we know that if its largest Jordan block is of size $n'$ then $(J^m-\lambda^m I)^{n'}$ will vanish and $(J^m-\lambda^m I)^{n'-1}$ won't).

As we saw above, the matrix $J^m$ is of the form $a_0I+a_1N+\dots+a_mN^m$, where the coefficients are given by $a_i=\binom mi\lambda^{m-i}$. Subtracting off $\lambda^mI$ gives $a_1N+\dots+a_mN^m$. So $(J^m-\lambda I)^k=(a_1N+\dots+a_mN^m)^k=b_kN^k+b_{k+1}N^{k+1}+\dots+b_{mk}N^{mk}$, where the coefficients $b_j$ can be worked out in terms of the $a_i$. As I said above, the matrix $N^j$ has ones on the $j$th diagonal, and zeros elsewhere. So the lowest order term $b_kN^k$ has non-zero entries on the $k$th diagonal, and all the other terms $b_jN^j$ have their non-zero entries on the $j$th diagonal, with $j>k$. So the $b_kN^k$ term is the only one affecting the $k$th diagonal, which means that if $b_kN^k$ is non-zero, then the whole expression $(J^m-\lambda I)^k$ is non-zero.

Working out $b_k$ we find it is equal to $m^k\lambda^{(m-1)k}$. Since $\lambda\neq 0$ (and the characteristic is non-zero) we can see that $b_k$ is non-zero. Which means that $b_kN^k$ is non-zero iff $N^k\neq 0$, i.e. iff $k<n$. So whenever $\lambda\neq 0$ the largest Jordan block in fact has size $n$, which means the Jordan form for $J^m$ is just the $n\times n$ block with eigenvalue $\lambda^m$.

Now consider the case where $\lambda=0$ (and so $J=N$). Note that if $(e_i)_{i=1,\dots ,n}$ is the standard basis, then we have $Ne_i=e_{i+1}$ for $i<n$ and $Ne_n=0$. In other words $N$ takes each basis vector to the next one, except for the last one, which it kills. Iterating $m$ times, we get that $N^me_i=e_{i+m}$ for $i\leq n-m$ and $N^me_i=0$ otherwise. Or in other words $N^m$ takes each basis vector to the one $m$ in front of it, except for the last $m$ which it kills.

Start with $e_1$ and repeatedly apply $N^m$. We get a sequence $e_1, e_{m+1},e_{2m+1}, \dots$ counting up in $m$s. Eventually we reach some maximal term $e_{km+1}$ after which the sequence dies, i.e. $N^me_{km+1}=0$. This means that $N^m$ acts on the vectors $(e_1, e_{m+1}, \dots,e_{km+1})$ as a Jordan block of size $k+1$. Similarly, $N^m$ acts as a Jordan block on every other sequence of vectors containing every $m$th vector in our original basis. We get $m$ such sequences starting at $e_1,\dots,e_{m}$. Each basis vector is contained in exactly one of these sequences. So we can write $N^m$ in Jordan normal form just by rearranging our basis into these $m$ sequences. What is the exact size of these Jordan blocks? Each sequence starts at some $e_i$ with $1\leq i\leq m$ and counts up in steps of $m$ finishing when it gets to the last entry which is $\leq n$. Suppose $m$ divides $q$ times into $n$ with a remainder of $r$. Then the sequences beginning with $e_i$ for $1\leq i\leq r$ will have length $q+1$ and the sequences beginning with $e_i$ for $r<i\leq m$ will have length $q$. So $J^m$ has $m$ Jordan blocks: $r$ of size $q+1$ and $m-r$ of size $q$.

• You should maybe clarify at the beginning that you are assuming characteristic $0$. – Eric Wofsey Jul 5 '16 at 16:29
• @EricWofsey Yeah. – Oscar Cunningham Jul 5 '16 at 16:36
• @OscarCunningham Could you explain the lowest order business in some more detail please? – linalg Jul 5 '16 at 21:33
• @OscarCunningham $N$ represents $e_i\mapsto e_{i-1}$. – Arrow Jul 5 '16 at 22:04
• @Arrow It depends on whether your convention puts the $1$s above the main diagonal or below it. The two conventions are equivalent by relabelling the basis so the indices run in the opposite order, i.e. setting $e'_i=e_{n+1-i}$. – Oscar Cunningham Jul 5 '16 at 22:10