Assume two commuting $n\times n$ complex matrices $A$ and $B$ are given. Then it is in general false that if $C$ is a square root of $A$, i.e., if $C^2=A$, then $C$ commutes with $B$ (the simplest counterexample is provided by $A$ the 2-by-two identity matrix and $C=diag(1,-1)$.

Yet, when $A$ is invertible, by using holomorphic functional calculus it is easy to show that there exist at least a square root of $A$ which commutes with $B$, i.e., there exist at least one $C$ such that $C^2=A$ and $[B,C]=0$.

Is there also an elementary (i.e., purely linear algebra) proof of this fact?


3 Answers 3


As in Geoff Robinson's answer, let us show that there exists a square root of $A$ which is a polynomial in $A$; then surely it commutes with $B$.

Let $\mu(x)$ be the minimal annihilating polynomial for $A$, $\mu(x)=\prod_i(x-\lambda_i)^{\alpha_i}$. We need to find a polynomial $P(x)$ such that $P(x)^2\equiv x\pmod {\mu(x)}$ (then $P(A)$ is a desired square root). By the Chinese remainder theorem, it suffices to find such polynomial modulo each of $(x-\lambda_i)^{\alpha_i}$, which is the same as finding a square root of $t+\lambda_i$ modulo $t^{\alpha_i}$. This last root can be proved to exist either by simple lifting of exponent $\alpha_i$, or by mentioning that the Taylor polynomial $$ P_i(t)=\sqrt{\lambda_i}\sum_{n=0}^{\alpha_i-1} \frac{\frac12\bigl(\frac12-1\bigr)\cdots\bigl(\frac12-n+1\bigr)}{n!}\biggl(\frac{t}{\lambda_i}\biggr)^n $$ for $(t+\lambda_i)^{1/2}$ fits.

This proof works as well if $\mu(x)$ is divisible by $x$ (but not by $x^2$) since $0$ is a square root of $x$ modulo $x$.

EDIT. As Geoff Robinson mentioned, my last statement was wrong, so the case when $\mu(x)$ is divisible by $x^2$ needs another methods. In this case, the desired claim is wrong. One may clearly use the statement that each operator commuting with all operators commuting with $A$ is a polynomial in $A$, but here is an explicit counterexample (surely inspired by Geoff).

Let $A=\pmatrix{0&0&1\\0&0&0\\0&0&0}$, $B=\pmatrix{1&0&0\\0&0&0\\0&0&1}$. Then $AB=BA$, and there exists a square root of $A$, for instance, $\pmatrix{0&1&0\\0&0&1\\0&0&0}$. On the other hand, since $B$ is the projector onto $\langle e_1,e_3\rangle$, each $C$ commuting with $B$ should have the form $C=\pmatrix{a&0&b\\0&c&0\\d&0&e}$. Now, if $C^2=A$ then $\pmatrix{a&b\\d&e}^2=\pmatrix{0&1\\0&0}$ which is impossible: otherwise we would obtain a $2\times 2$ matrix whose square is nonzero but the fourth power is zero.

  • $\begingroup$ I don't think the Very last statement is necessarily true. In a 3 dimensional space a nilpotent Matrix with min polynomial of degree 2 has a square root. $\endgroup$ Apr 1, 2014 at 17:52
  • $\begingroup$ @GeoffRobinson: Oh yes, ypu are right. Sorry. I have added some words about this case; hopefully now they are correct;)... $\endgroup$ Apr 1, 2014 at 18:44
  • $\begingroup$ A similar question: Suppose that $A$, $B$, $C$ are real matrices and they are pair-wise commuting. Does there exists a real matrix $R$ such that $R^2=A^2+B^2$ and $R$ commutes with $C$? I know that there exists a real matrix $R$ such that $R^2=A^2+B^2$ but I do not see an easy way to prove the commutativity without invoking some structure theorem like real Schur form. Specifically, I'd like to ask, does $A^2+B^2$ has a real square root that is a polynomial in both $A$ and $B$? (I know that it's sometimes impossible to be a polynomial in $A^2+B^2$) $\endgroup$
    – user58955
    Sep 26, 2014 at 15:13

Here is a linear algebra proof, expanding my deleted comment, )which was OK in the invertible case), to discuss what happens in the non-invertible case.: we may write $A = S + N,$ where $S$ is diagonalizable, $N$ is nilpotent, and both $S$ and $N$ are polynomials in $A.$ Suppose first that $S$ is invertible. Then $S$ has a square root which is a polynomial in $S$ ( and hence also in $A$), using Lagrange interpolation). Hence we need to construct a square root of $I + S^{-1}N$ which is still a polynomial in $A.$ Note that $S^{-1}$ is also a polynomial in $A.$ Now the expansion of $(I +S^{-1}N)^{\frac{1}{2}}$ by the binomial theorem is a finite sum, since $S^{-1}N$ is nilpotent, and is hence expressible as a polynomial in $A.$ If $S$ is not invertible, but $N$ annihilates the null space of $S,$ then this argument can be adjusted to give the same conclusion. However, if $S$ is not invertible, but $N$ does not annihilate the null space of $S,$ then $A$ need not have a square root ( eg if $S = 0$ and $N$ acts as a single Jordan block of maximal size). If $A$ does have a square root $C,$ then $C$ commutes with both $S$ and $N,$ and $C$ leaves the null space of $S$ (and the sum of the non-zero eigenspaces of $S$) invariant. We may write $C = T + M,$ where $T,M$ are polynomials in $C$, $T$ is diagonalizable, and $M$ is nilpotent. Then $M$ and $T$ each commute with both $S$ and $N.$ Both $T$ annihilates the null space of $S$ and $M^{2}$ and $N$ each act the same way on the null space of $S.$ It follows that $A$ may have such a square root under certain conditions on the size of Jordan blocks. However, such a square root need not be a polynomial in $A.$


Let $A$ be an invertible $n\times n$ matrix. Without loss of generality, we may assume that the spectrum of $A$ does not meet the half-line $(-\infty,0]$: in fact, it is possible to find a half-line with vertex 0 which does not meet the spectrum of $A$. We define then $$ \text{Log}\ A=\oint_{[I,A]}\frac{d\xi}{\xi}=\int_0 ^1(A-I)\bigl(I+t(A-I)\bigr)^{-1}dt, \tag {$\ast$} $$ which makes sense since for $t\in (0,1]$, $$ x+t(A-I)x=0\Longrightarrow Ax=-(1-t)t^{-1}x\Longrightarrow \sigma(A)\cap\mathbb R_-\not=\emptyset. $$ Using an analytic continuation argument, we get that $\exp(\text{Log}\ A)=A$ and we define $$ A^{1/2}=\exp(\frac12\text{Log}\ A),\quad\text{which is indeed such that $(A^{1/2})^2=A$.} $$

Now if $B$ is a $n\times n$ matrix commuting with $A$, it commutes with $\text{Log}\ A$, as it can be seen from $(\ast)$ and thus with the $A^{1/2}$ defined above.

  • $\begingroup$ Right, this is the holomorphic functional calculus argument I was referring to. $\endgroup$ Apr 1, 2014 at 16:04
  • $\begingroup$ Yes, but just a remark here: this holomorphic method provides an explicit formula that can be numerically calculated and approximated. The algebraic proofs above require the knowledge of unknown quantities, such as the minimal polynomial, or the Jordan form. Although perfect theoretically, an algebraic method for this problem will require a very large time to provide a simple approximation of a commuting square-root: the numerical cost of the determination of the minimal polynomial or of the Jordan form is huge, compared to the simple algorithm to approximate a simple integral. $\endgroup$
    – Bazin
    Apr 1, 2014 at 19:53

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