Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Under what conditions does a matrix $A$ have a square root? I saw somewhere that this is true for Hermitian positive definite matrices(whose definition I just looked up).

Moreover, is it possible that for some subspace $X \subset M_n(\mathbb R)$ of $n\times n$ matrices over $\mathbb R$, the map $A \mapsto \sqrt{A}$ is continuous? People who want to consider more generality can also look at matrices over $\mathbb C$.

Thank you.

share|improve this question
5  
A related question. If your matrix possesses a Jordan block with a zero eigenvalue, you're shot. In general you need to peer at the Jordan form of your matrix and see if the square root function (or any other matrix function for that matter) is defined on your Jordan blocks. –  J. M. Sep 17 '11 at 6:15
    
@J.M.: Is looking at the Jordan form a continuous operation? –  Derek Sep 17 '11 at 6:35
4  
@J.M. That is not true. For all nilpotent Jordan blocks larger than $2\times 2$, squaring them yields a nontrivial endomorphism. This generates tons of nilpotent matrices which have square roots, even though no individual Jordan block has a square root. There is, in fact, a nice combinatorial description of the nilpotent matrices which have square roots, and which gives the Jordan form of all possible square roots (in general, there are several). Namely, if $a_i=\dim \ker A^i-\dim \ker A^{i-1}$, then $A$ will have a square root iff $a_i=a_{i+1}\Rightarrow a_i$ is even. –  Aaron Sep 20 '11 at 16:38
    
I suppose you're right, @Aaron, but would you happen to have an explicit example of a "rootable" nilpotent on hand, just so I can check my intuition? :) –  J. M. Sep 20 '11 at 16:46
1  
@J.M. Consider the $4\times 4$ matrix with two $2\times 2$ blocks. This is the square of a matrix conjugate to a $4\times 4$ Jordan block, unless I've done my calculations wrong. –  Aaron Sep 20 '11 at 16:49

4 Answers 4

Some partial answers to your question can be derived from this answer. Here are two examples:

(1) If $A\in M_n(\mathbb C)$ is invertible, then there is an open neighborhood $U$ of $A$ in $M_n(\mathbb C)$ and a holomorphic function $f:U\to M_n(\mathbb C)$ such that $f(B)^2=B$ and $f(B)\in\mathbb C[B]$ for all $B$ in $U$.

(2) Let $U$ be the set of all $A$ in $M_n(\mathbb C)$ such that no eigenvalue of $A$ is a nonpositive real number. Then $U$ is open in $M_n(\mathbb C)$. Moreover, there is a holomorphic function $f:U\to M_n(\mathbb C)$ such that we have for all $A$ in $U$:

  • $f(A)^2=A$,

  • $f(A)\in\mathbb C[A]$,

  • if $A$ is a positive definite hermitian matrix, then $f(A)$ is the usual square root of $A$.

(Marc van Leeuwen first comment below refers to a former version of the answer.)

share|improve this answer
    
Fair enough. But I don't really see which part of the question this answers. –  Marc van Leeuwen Apr 4 '13 at 5:41
    
Dear Marc: Thanks for your comment. I'll answer it as soon as possible, but I'll be very busy today and tomorrow. I'll try to get back to you Saturday (Paris time) at the latest. –  Pierre-Yves Gaillard Apr 4 '13 at 12:46
    
Dear Marc: I've just edited the answer. Thank you again for your interest! –  Pierre-Yves Gaillard Apr 6 '13 at 7:39

Have a look at the wikipedia article http://en.wikipedia.org/wiki/Matrix_square_root

which I found exceptionally good ---- much better than average for Wikipedia!

share|improve this answer

Over the complex numbers (or any other algebraically closed field with $\operatorname{char} k\neq 2$), every invertible matrix has a square root. In fact, over $\mathbb C$, since every invertible matrix has a logrithm, we can take a one parameter family of matrices $e^{t\log A}$, and taking $t=1/2$ yields a square root of $A$. To see the existance of matrix logrithms, it suffices to show that $I+N$ has a logrithm, where $N$ is nilpotent, and this follows from Taylor series (similar to Ted's proof of the existence of sqare roots).

Thus, we can determine if a matrix $A$ has a square root by restricting to $\displaystyle\bigcup_n \ker A^n$, which is the largest subspace on which $A$ acts nilpotently. In what follows, we will assume that $A$ is nilpotent.

Up to conjugation, $A$ is determined by its Jordan normal form. However, equivalent to JNF for a nilpotent matrix is the data $a_i'=\dim \ker A^i$ for all $i$. This is obviously an increasing sequence. Less obvious is that the sequence $(a_i)$ where $a_i=a'_i-a'_{i-1}$ is a decreasing sequence, and hence forms a partition of $\dim V$ where $A:V\to V$. We note that this data is equivalent to the data in JNF, as $a_i-a_{i+1}$ will be the number of Jordan blocks of size $i$. More explicitly, a jordan block of size $k$ corresponds to the partition $(1,1,1,1,1\ldots, 0,0,0,\ldots)$ with $k$ $1's$, and if a nilpotent matrix $A=\oplus A_i$ is written in block form where each block $A_i$ corresponds to a partition $\pi_i$, then $A$ corresponds to the partition $\pi=\sum \pi_i$, where the sum is taken termwise, e.g. $(2,1)+(1,1)+(7,4,2)=(10,6,2)$.

Moreover, $A^2$ corresponds to the partition $(a_1+a_2, a_3+a_4,\ldots, a_{2i-1}+a_{2i}, \ldots).$ Because every matrix will be conjugate to a JNF matrix and $\sqrt{SAS^{-1}}=S\sqrt{A}S^{-1}$, we see that a matrix will have a square root if and only if the corresponding partition has a "square root."

The only obstruction to a partition having a square root is if two consecutive odd entries are equal. Otherwise, we can take one (of many) square roots by replacing each $a_i$ with the pair $\lceil a_i/2 \rceil, \lfloor a_i/2 \rfloor$.

share|improve this answer

Expanding on J.M.'s comment, if we look at a Jordan block $J$ of size $n$ with eigenvalue 0, then I claim that if $n>1$, $J$ has no square root. Suppose $K^2 = J$. Then $K^{2n} = J^n = 0$, but $K$ is an $n$ by $n$ matrix, so in fact $K^n=0$. If $n$ is even, we conclude that $J^{n/2}=0$ (since $K^2=J$), a contradiction since no power of $J$ less than $n$ can be 0. If $n$ is odd, then we conclude $J^{(n-1)/2} K = 0$, and multiplying $K$ on the right gives $J^{(n+1)/2} = 0$. If $n>1$ this is again a contradiction.

On the other hand, if we look at a Jordan block $J$ with eigenvalue $\lambda \ne 0$, then we may write $J = \lambda I + N$ where $N$ is a nilpotent matrix. To find a square root of $J$, expand $(\lambda I + N)^{1/2}$ using the usual Taylor series for the square root around the point $\lambda$, with an increment of $N$. Since $\lambda \ne 0$, all derivatives of the square root are defined at $\lambda$ and since $N$ is nilpotent, the Taylor series terminates and no convergence issues arise.

share|improve this answer
1  
Although a single non-trivial nilpotent Jordan block does not have a square (or higher) root, this does not settle the question for nilpotent matrices. If you've got a $3\times3$ nilpotent matrix with Jordan blocks of sizes $(2,1)$, then it does have a square root. Just compute (the Jordan normal form of) the square of a Jordan block of size $3$ to see this. In general you need to consider the whole multiset of nilpotent Jordan block sizes to be able to decide. See this answer. –  Marc van Leeuwen Apr 1 '13 at 10:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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