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Square roots of nilpotent matrices December 2, 2008

Posted by Ben Webster in linear algebra.
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Oddly enough, it seems that after all the discussion at this post, no-one actually wrote down a criterion for a nilpotent matrix to have a square root.

The answer is very nice, in fact:

Proposition. A nilpotent matrix has a square root if and only if the (2n-1)-th Jordan block (from largest to smallest) is the same size, or one larger than 2n-th.

Who feels like giving a proof in comments?

EDIT: I originally bollixed the condition, by transposing the partition given by the Jordan blocks (the above condition is the same as the partition given by the Jordan blocks having transpose where each odd row length appears at most once).

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Comments

1. Noah Snyder - December 2, 2008

I think you’re wrong… Isn’t the square of a nilpotent 6-block, two 3-blocks? Also (1,1,1)^2 = (1,1,1).

2. Ben Webster - December 2, 2008

Wrong is such an ugly word. I just got my partitions transposed. Now your counterexample has been itself countered.

3. Noah Snyder - December 2, 2008

Still wrong! Simplest counterexample is (2,3,4).

4. Scott Carnahan - December 2, 2008

If I’m not mistaken, the square of a nilpotent 2n-block is two n-blocks, while the square of a (2n+1)-block is an n-block and an (n+1)-block. You can then order the blocks by decreasing size, separate into pairs, and note that admissible pairs have size off by at most one. We can eliminate the boundary case of an odd number of blocks by allowing blocks of size zero.

5. Noah Snyder - December 2, 2008

I agree with Scott!

6. Ben Webster - December 2, 2008

Still wrong! Simplest counterexample is (2,3,4).

Not a counterexample. The 4th part is 0 and 3rd part is 2.

Good proof, Scott.

7. Noah Snyder - December 2, 2008

Ah, ok, I misunderstood your new statement. Good thing I thought better of my plan to scratch out this posts title and replace it with “Epic Fail.”

8. Ben Webster - December 2, 2008

Ex-squeeze me? They make lolposts now?

9. Greg Kuperberg - December 2, 2008

I did give the condition in comment #17: “The nilpotent term has a square root iff Jordan blocks can be paired so that the two blocks in each pair are either the same size or adjacent sizes.”

The proof follows directly from the Jordan canonical form of the square root.

10. Scott Carnahan - December 2, 2008

im in ur matrix killing squaring ur bl0x

11. Square Roots « The Unapologetic Mathematician - March 10, 2009

[...] Over at the Secret Blogging Seminar, they discussed this without using the Jordan normal form, and they dealt with the nilpotent blocks later, which has a similar feel to what we’re [...]


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