9
$\begingroup$

Let $A$ be a $n\times n$ matrix with characteristic polynomial $$(x-c_{1})^{{d}_{1}}(x-c_{2})^{{d}_{2}}...(x-c_{k})^{{d}_{k}}$$ where $c_{1},c_{2},...,c_{k}$ are distinct. Let $V$ be the space of $n\times n$ matrics $B$ such that $AB=BA$. How to find dimension of this vector space? Clearly it is easy to find dimension if the matrix $A$ is given diagonalizable but how to find dimension if matrix $A$ is not diagonalizable. I tried it by using Jordan canonical form but its very lengthy and only gives possible dimensions. Can some one suggest how to find by giving a particular matrix.

$\endgroup$
6
  • $\begingroup$ There is no single answer given the information you have. There is one kind of extreme, which is when only matrices that can be written as a polynomial $p(A)$ commute with $A.$ This has small dimension, just $n.$ I collected a bunch of equivalent conditions in my answer at math.stackexchange.com/questions/92480/… The other extreme is the identity matrix, everything commutes with it, dimension $n^2,$ much, much larger. $\endgroup$ – Will Jagy May 21 '15 at 3:07
  • $\begingroup$ It's not enough to know the characteristic polynomial. You really have to know the size of the Jordan blocks. $\endgroup$ – user148177 May 21 '15 at 3:09
  • $\begingroup$ yes dimension is atleast n....but if it not diagonalizable then is there some othere technique to find dimension? $\endgroup$ – neelkanth May 21 '15 at 3:10
  • $\begingroup$ @WillJagy if the exact size of each block is known then we look dimension of commutators of each blocks and add?? $\endgroup$ – neelkanth May 21 '15 at 3:15
  • $\begingroup$ I think you can only do that if the blocks have different eigenvalues. Otherwise it's more complicated. $\endgroup$ – user148177 May 21 '15 at 3:17
2
$\begingroup$

Here is an example with two Jordan blocks that have the same eigenvalue. If we had only a single Jordan block, the dimension resulting would be exactly $4,$ but this comes out a little bigger.

$$ \left( \begin{array}{rr|rr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right) = \left( \begin{array}{rrrr} e & f & g & h \\ 0 & 0 & 0 & 0 \\ m & n & o & p \\ 0 & 0 & 0 & 0 \end{array} \right) $$

$$ \left( \begin{array}{rrrr} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right) \left( \begin{array}{rr|rr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) = \left( \begin{array}{rrrr} 0 & a & 0 & c \\ 0 & e & 0 & g \\ 0 & i & 0 & k \\ 0 & m & 0 & o \end{array} \right) $$

At first glance, I get eight linear equations required for equality, $$e,g,m,o = 0, \; a=f, c=h, i=n,k=p. $$

Out of $16,$ what is left is dimension $8.$

$$ \left( \begin{array}{rr|rr} a & b & c & d \\ 0 & a & 0 & c \\ \hline i & j & k & l \\ 0 & i & 0 & k \end{array} \right) $$ Dimension $8$ for four little 2 by 2 Jordan blocks scattered about.

$\endgroup$
6
  • $\begingroup$ OK WE CAN TRY THIS FOR EACH BLOCK... $\endgroup$ – neelkanth May 21 '15 at 3:30
  • $\begingroup$ There's a typo. The last column of the second equation on the right hand side should be $c, g, k, o$, and the dimension is 8 (throw out $l = 0$ ) $\endgroup$ – user148177 May 21 '15 at 3:43
  • $\begingroup$ Also, this shows you can't just find the centralizers for each block and then add dimensions. Note that the upper $2 \times 2$ matrix is not forced to be zero by the equations. So it's more complicated. $\endgroup$ – user148177 May 21 '15 at 3:46
  • $\begingroup$ @user148177, yes, i went through it again, had a letter wrong, this is much better. I try to get the students to experiment; with matrices, if the problem says $n$ by $n,$ see what happens with $2$ by $2$ or $3$ by $3.$ Usually they just don't. I did $4$ by $4$ here because I wanted very clear distinct Jordan blocks with the little 1's off diagonal. $\endgroup$ – Will Jagy May 21 '15 at 3:51
  • 1
    $\begingroup$ @WillJagy Yeah my second comment is mostly in reply to a comment to the question where the asker suggested just computing centralizers of the blocks and adding dimensions, which I think is wrong in general. Anyway, the only way I know how to nicely understand this stuff in general uses some $\mathfrak{sl}_2$ representation theory, which I think is maybe a little beyond the scope here... $\endgroup$ – user148177 May 21 '15 at 3:54
3
$\begingroup$

And, in case you were wondering, here's the case for two Jordan blocks of sizes $3$ and $2$. The centralizer of $$ \left[ \begin {array}{ccccc} \lambda&1&0&0&0\\0&\lambda&1&0&0 \\ 0&0&\lambda&0&0\\ 0&0&0&\lambda&1 \\ 0&0&0&0&\lambda\end {array} \right] $$ consists of matrices of the form $$ \left[ \begin {array}{ccccc} b_{{3,3}}&b_{{2,3}}&b_{{1,3}}&b_{{2,5}}& b_{{1,5}}\\ 0&b_{{3,3}}&b_{{2,3}}&0&b_{{2,5}} \\ 0&0&b_{{3,3}}&0&0\\0&b_{{5,3}} &b_{{4,3}}&b_{{5,5}}&b_{{4,5}}\\0&0&b_{{5,3}}&0&b_{ {5,5}}\end {array} \right] $$ More generally, if you have Jordan blocks with the same eigenvalue for indices $i, \ldots, j$ and $k, \ldots, l$, for a matrix in the centralizer the block formed by rows $i,\ldots, j$ and columns $k, \ldots, l$ will be "upper triangular" (i.e. $0$ for $row - column > \min(i-k, j-l)$) and constant on diagonals. The dimension for this block is then $\min(l-k+1,j-i+1)$, and you have to add this up for all blocks.

$\endgroup$
3
$\begingroup$

I don't think there's an easy way to do this. I agree with the above comment that you should play with different matrices yourself. I'm adding this mostly so you have a reference. See: Nilpotent Orbits in Semisimple Lie Algebras by Collingwood, McGovern, Theorem 6.1.3. The statement is:

Let $A$ be a nilpotent matrix, i.e. eigenvalues are all zero. Its Jordan normal form is given by a partition of $n$, say $d_1 + \cdots + d_k$. Then, one has that the dimension of the centralizer is $$\sum s_i^2$$ where $s_i = |\{j \mid d_j \geq i\}|$. You can generalize this formula now to a general matrix $A$ in Jordan normal form (the blocks with different eigenvalues don't interact -- you can check quickly if you think of $A$ as acting by column/row operations), to get $$\sum_\lambda \sum_i s_{i, \lambda}^2$$ where $\lambda$ ranges over all eigenvalues (generalized) and $s_{i, \lambda}$ as above.

Edit for experts: Okay, so, the way I would do this in general, is for a nilpotent element $x$, find $h$ such that $[h, x] = 2x$. One can do this separately for each Jordan block; if $x$ is a Jordan block, then $h = diag(k, k-2, k-4, \ldots, 4-k, 2-k, -k)$. One can complete this to an $\mathfrak{sl}_2$ triple by taking $y = x^t$, i.e. $(x, y, h)$ is a copy of $\mathfrak{sl}_2$. Then, we can decompose $\mathfrak{g}$ into into eigenspaces for $ad(h)$, and note that in fact these are eigenspaces for various $\mathfrak{sl}_2$-representations, and the highest weights are killed by $x$ (i.e. in the centralizers). So, we want to count representations, which one can do with some combinatorics (yielding the above formula) or just directly. For example, in the above (or below) answer, $h = diag(1,-1,1,-1)$, and one has $$\left(\begin{array}{cccc}0&2&0&2\\-2&0&-2&0\\0&2&0&2\\-2&0&-2&0\end{array}\right)$$ where the number is the $h$-weight of that eigenspace. We count 4 weight 2 eigenspaces, 4 weight -2 eigenspaces, and 8 weight 0 eigenspaces, giving us a total of $4 + (8-4) = 8$ representations. Explicitly, one has the decomposition $\mathfrak{g} = V(2)^{\oplus 4} \oplus V(0)^{\oplus 4}$ as $\mathfrak{sl}_2$ representations.

$\endgroup$
2
  • $\begingroup$ its solves atleast some particular cases... $\endgroup$ – neelkanth May 21 '15 at 3:35
  • $\begingroup$ If I'm not mistaken I think this gives all cases $\endgroup$ – user148177 May 21 '15 at 3:44

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.