5
$\begingroup$

Let $J$ be a $k \times k$ jordan block, prove that any matrix which commutes with $J$ is a polynomial in $J$.

I appreciate your hints, Thanks

$\endgroup$
6
  • 1
    $\begingroup$ quite a distance to the full result, which is more than you asked. Do the 2 by 2 and 3 by 3 cases by hand, you will learn something. Actually, the 1 by 1 case as well... $\endgroup$ – Will Jagy Aug 18 '14 at 19:03
  • $\begingroup$ Thanks, I can prove (doing case by case) that All powers of $J$ commute with $J$, but that does not prove anything... should I assume that some $2 \times 2$ matrix which is not a polynomial in $J$ does not commute with $J$ ? $\endgroup$ – the8thone Aug 18 '14 at 19:08
  • $\begingroup$ You will do what you want, I suppose. I suggest you write out a 2 by 2 Jordan block, call the eigenvalue e or something, write a general 2 by 2 matrix with entries a,b,c,d, multiply in both orders and see what conditions make them commute. Because of Cayley-Hamilton, for 2 by 2 any polynomial need only be linear, $M = A I + B J$ $\endgroup$ – Will Jagy Aug 18 '14 at 19:17
  • $\begingroup$ Let's see, for 3 by 3 9 entries, maybe a,b,c,d,e,f,g,h,i, one eigenvalue, maybe w, and the polynomials to be considered need only be quadratic, $AI + B J + C J^2.$ Anyway, I think you are a bit over your head, and hands-on manipulation of concrete examples, small enough to be done by a human being, is going to teach you more than me parroting some proof. The very best outcome is if the concrete examples lead you to your own proof. $\endgroup$ – Will Jagy Aug 18 '14 at 19:24
  • 1
    $\begingroup$ Another easy way to prove the result (well depending on how much theory you know) is the fact that each Jordan block necessarily has a cyclic vector since the minimal and characteristic polynomials for a Jordan block are equal. This in turn implies (via the cyclic vector theorem) that each matrix which commutes with the $J$ is a polynomial in $J$. $\endgroup$ – EuYu Aug 18 '14 at 20:11
7
$\begingroup$

One direction of the equivalence is easy. For any polynomial $p$ and any square matrix $A$, $p(A)A=Ap(A)$.

For the other direction, we can assume that the $k\times k$ Jordan block $J_k$ has zeros on the diagonal. Indeed, any other Jordan block can be written in the form $\beta I + J_k$ and it is easy to see that $A$ commutes with $\beta I+J_k$ if and only if it commutes with $J_k$.

Let $A=(a_{ij})$ be a $k\times k$ matrix and assume that $AJ_k-J_kA=0$. We first show that the this implies that $A$ is an upper triangular (UT) and Toeplitz (T) matrix and proceed by induction on $k$. So assume that if a $(k-1)\times (k-1)$ matrix commutes with a $(k-1)\times (k-1)$ Jordan block, it is UT&T. Let's write $J_k$ and $A$ in the partitioned form $$ A=\begin{bmatrix}A_{11}&a_{12}\\a_{21}^*&\alpha_{22}\end{bmatrix}, \quad J_k=\begin{bmatrix}J_{k-1}&e_{k-1}\\0&0\end{bmatrix}, $$ where $A_{11}$ and $J_{k-1}$ are $(k-1)\times(k-1)$. Let's have a look on the commutator: $$ \begin{split} AJ_k-J_kA&=\begin{bmatrix}A_{11}&a_{12}\\a_{21}^*&\alpha_{22}\end{bmatrix}\begin{bmatrix}J_{k-1}&e_{k-1}\\0&0\end{bmatrix}-\begin{bmatrix}J_{k-1}&e_{k-1}\\0&0\end{bmatrix}\begin{bmatrix}A_{11}&a_{12}\\a_{21}^*&\alpha_{22}\end{bmatrix}\\ &=\begin{bmatrix}A_{11}J_{k-1}-J_{k-1}A_{11}-e_{k-1}a_{21}^*&A_{11}e_{k-1}-J_{k-1}a_{12}-\alpha_{22}e_{k-1}\\a_{21}^*J_{k-1}&a_{21}^*e_{k-1} \end{bmatrix}\\ &=\begin{bmatrix}A_{11}J_{k-1}-J_{k-1}A_{11}-e_{k-1}a_{21}^*&(A_{11}-\alpha_{22}I)e_{k-1}-J_{k-1}a_{12}\\a_{21}^*J_{k-1}&a_{21}^*e_{k-1} \end{bmatrix}. \end{split} $$

Since $AJ_k-J_kA=0$ by assumption, the last two block rows imply that $$ a_{21}^*e_{k-1}=0, \quad a_{21}^*J_{k-1}=0. $$ The first equation implies that the last entry of $a_{21}=0$ is zero, while the other says that the first $k-2$ entries of $a_{21}$ are zero. Therefore, $a_{21}=0$ and indeed the matrix $A$ is UT (since by the induction assumption, $A_{11}$ is UT).

The first block of $AJ-JA=0$ implies that $$ A_{11}J_{k-1}-J_{k-1}A_{11}-e_{k-1}a_{21}^*=A_{11}J_{k-1}-J_{k-1}A_{11}=0 $$ because we already showed that $a_{21}=0$ and hence from $A_{11}J_{k-1}-J_{k-1}A_{11}=0$ we have that $A_{11}$ is UT&T (by the induction assumption).

It remains to show that if $A_{11}$ is T then from $(A_{11}-\alpha_{22}I)e_{k-1}-J_{k-1}a_{12}=0$ we have that $A$ is T. We have $$ (A_{11}-\alpha_{22}I)e_{k-1} = \begin{bmatrix} a_{1,k-1}\\ \vdots\\ a_{k-2,k-1}\\ a_{k-1,k-1}-\alpha_{22} \end{bmatrix}, \quad J_{k-1}a_{12}=\begin{bmatrix} a_{2,k}\\ \vdots\\ a_{k-1,k}\\ 0 \end{bmatrix}. $$ Since both vectors are equal, it gives $a_{i,k-1}=a_{i+1,k}$ for $i=1,\ldots,k-2$ and $\alpha_22=a_{k-1,k-1}$. Since $A_{11}$ is UT&T, the diagonal of $A_{11}$ is constant and thus the diagonal of $A$ is constant as well and the last $k-2$ entries of $a_{12}$ are "copies" of the first $k-2$ entries of the last column of $A_{11}$. Therefore, $A$ is T.

Summarizing, $A$ and $J_k$ commute implies that $A$ is a UT&T matrix and consequently it can be written in the form $$ A=\begin{bmatrix} a_1 & a_2 & \ldots & a_k \\ & a_1 & \ldots & a_{k-1} \\ & & \ddots & \vdots \\ & & & a_1 \end{bmatrix}. $$ By realizing how the powers of $J_k$ look like, it is easy to see that $A$ can be written as $$ A=a_1 J_k^0 + a_2 J_k^1 + \ldots + a_k J_k^{k-1} =: p(J_k), \quad p(t)=a_1+a_2t +\ldots+a_k t^{k-1}. $$

Q.E.D.

$\endgroup$
1
$\begingroup$

2 by 2 only. i really hope you will try the 3 by 3 case by hand, analogous to this:

Aright, eigenvalue $w,$ $$ J = \left( \begin{array}{rr} w & 1 \\ 0 & w \end{array} \right), $$ trial $$ M = \left( \begin{array}{rr} a & b \\ c & d \end{array} \right), $$

Next $$ JM = \left( \begin{array}{rr} wa + c & wb+d \\ wc & wd \end{array} \right), $$ but $$ MJ = \left( \begin{array}{rr} wa & wb+a \\ wc & wd+ c \end{array} \right). $$

We find that $JM = MJ$ precisely when $$ c = 0 \; \; \mbox{AND} \; \; a=d. $$ Notice that $w$ does not appear. Under these conditions, $$ M = \left( \begin{array}{rr} a & b \\ 0 & a \end{array} \right). $$ That means that $$ M = (a-bw)I + b J, $$ that is a polynomial in $J.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.