2
$\begingroup$

I have a question concerning the definition of the square root of bounded linear operators. To introduce some notation: tr denotes the trace of linear operators and $\mathcal{L}(H)$ denotes the set of bounded linear operators, from H to H, where H symbolizes a Hilbert space. L' stands for the adjoint operator of L. We introduced the the space of trace class operators in the following way: $\{L \in \mathcal{L}(H) \ : \ tr((LL')^{\frac{1}{2}} < \infty \}$. The problem is, that I don't know how how square root of the above mentioned operator is defined.

$\endgroup$
2
$\begingroup$

Notice that $LL'$ is positive definite. Square root is well-defined for bounded, positive definite operators (by functional calculus theorems).

$\endgroup$
  • $\begingroup$ I only read, what you have stated, but I can't actually imagine, what a quare root in the context of Hilbert spaces means, e.g. how to 'calculate' $Q^{\frac{1}{2}} h$ for an $h \in H$. $\endgroup$ – David Jan 11 '13 at 13:37
  • 1
    $\begingroup$ @David: for $\lvert x\rvert\leq 1$ we have $\sqrt{1-x}=1+\sum_{n=1}^\infty \binom {1/2} n (-x)^n$ (and the convergence is uniform). Using this series expansion you can define square root for positive operators $Q$ of norm not greater than $1$, and for others you can put $\sqrt Q= \sqrt {Q/\lVert Q\rVert }\cdot \sqrt {\lVert Q\rVert}$. For more detailed discussion, I suggest you check some functional analysis textbook. It should be explained in detail in proximity of polar decomposition of bounded operators. $\endgroup$ – tomasz Jan 11 '13 at 13:57
2
$\begingroup$

It is a well-known fact that every bounded positive operator on a Hilbert space (which are exactly those of the form $LL'$ for some bounded $L$) has a unique positive square root. This is a consequence of the continuous functional calculus; you can find about in any Functional Analysis book that talks about C$^*$-algebras.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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