0

Prove if a square root of a Herimitian positive semi-definite matrix is also Herimitian positive semi-definite, then it is unique.

Here is what I have done so far:

By the spectrum theorem, suppose $A$ and $B$ are Herimitian positive semi-definite matrices, then $A$ and $B$ can be written as $$A=Q_1D_1Q_1^*$$ $$B=Q_2D_2Q_2^*,$$ where $D_1, D_2$ are nonnegative diagonal matrices. Suppose $B^2=A$, by the similarity transformation, $D_1$~$D_2^2$.

How can I show $B$ is unique from here?

DDaren
  • 421
  • Hmm. Your argument for $D_1 = D_2^*$ is wonky – user251257 Oct 07 '15 at 12:06
  • I think that the identity is Hermitian positive semi-definite, and since we have $I^2 = I$, we have a square root that's also HPSD...but we also have $(-I)^2 = I$, so it's not unique. I'm probably missing something here, but this makes me doubtful. (If you complain that it's really positive DEFINITE, then add one more row and column of zeros, and you get one that PSD). – John Hughes Oct 07 '15 at 12:10
  • @user251257 Use $A=B^2$, move all unitary matrices to one side gives a similarity transformation $D_1$~$D_2^2$. Since they are diagonal matrix, $D_1=D_2^2$. – DDaren Oct 07 '15 at 12:10
  • @JohnHughes $-I$ is not positive definite... – DDaren Oct 07 '15 at 12:12
  • @DDaren $diag(4,1) \sim diag(1, 2)^2$ – user251257 Oct 07 '15 at 12:15
  • Ah: the problem means to say "then it is the unique PSD square root", not "it is the unique square root". The phrase "it is unique" is a little ambiguous here, although clearly my interpretation can't be right, because of the $\pm I$ example...but my pre-coffee mind couldn't manage to see the other possibility. – John Hughes Oct 07 '15 at 12:15
  • @user251257 you are right my argument is wrong. – DDaren Oct 07 '15 at 12:25
  • 1
    You need to argue why you may choose $Q_1 = Q_2$. hint: $A$ and $B$ commute. – user251257 Oct 07 '15 at 12:27
  • @user251257 Perhaps I should say that $A$ and $B$ have the same set of eigenvectors and just let $Q_2=Q_1$? – DDaren Oct 07 '15 at 12:42
  • @DDaren yes ${}{}$ – user251257 Oct 07 '15 at 12:46
  • @user251257 thank you – DDaren Oct 07 '15 at 12:54

0 Answers0