Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $A$ be a Hermitian positive semi-definite $n$ by $n$ matrix (The field of scalars is $\mathbb{C}$). Let $B$ be an $n$ by $n$ matrix that commutes with A. Prove that $B$ and $\sqrt{A}$ commute.

I started to solve the problem like this: Since $A$ is a Hermitian positive semi-definite matrix, then there exits a unitary matrix $J$ such that: $$J^{-1}AJ=D=diag(\lambda _{1},\lambda _{2},...,\lambda _{n})$$ where: $\lambda _{i}\geq 0$ and $\lambda _{i}$ are eigenvalues of $A$, and since $A$ is Hermitian, then the eigenvalues are real.

Let $$F=\sqrt{A}=JCJ^{-1}$$ where $C=\sqrt{D}=diag\left ( \sqrt{\lambda _{1}},\sqrt{\lambda _{2}},...,\sqrt{\lambda _{n}} \right )$, and $\sqrt{\lambda _{i}}$ are eigenvalues of $F=\sqrt{A}$.

$AB=BA\Rightarrow JDJ^{-1}B=BJDJ^{-1}$

So, I need to use the above equality to prove this one: $$FB=BF\Rightarrow JCJ^{-1}B=BJCJ^{-1}\Rightarrow ...$$

Please let me know how I should finish my proof. Also, if anyone has a different solution, please share. Thanks

share|cite|improve this question
+1 for the title typo: "semi-fefinite" :) – user2468 Feb 27 '12 at 22:50
up vote 3 down vote accepted

Not sure if my way would be very "linear-algebraic standard", but this is what feels natural to me:

Note that for any polynomial $p\in\mathbb{R}[x]$, $p(A)=J^{-1}p(D)J$ (just do it first for monomials, and see that it works for linear combinations of monomials).

Now observe that, since $BA=AB$, then $BA^2=BAA=ABA=AAB=A^2B$, etc., so we conclude that $Bp(A)=p(A)B$ for any polynomial $p$.

Finally, choose a polynomial $p$ such that $p(\lambda_j)=\sqrt{\lambda_j}$, $j=1,\ldots,n$. Then $p(D)=C$, in the notation from the question, which implies that $p(A)={A}^{1/2}$.

So $$ BA^{1/2}=Bp(A)=p(A)B=A^{1/2}B $$

share|cite|improve this answer
For all: Please let me know if anyone of you has another method of solving the problem. – M.Krov Feb 28 '12 at 5:34
Hint: essentially, two matrices commute iff they have the same eigenvectors. The square root of a matrix commutes w/the original matrix... – DVD Jun 12 '14 at 20:20

Your Answer


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.