2
$\begingroup$

Note that the vector space V is finite-dimensional.

So basically, we need to rewrite $T\in\mathcal{L}(V)$ as $T=a(S_1+S_2)$.I thought of using polar decomposition to write $T=G\sqrt{T^*T}$, but I'm not sure how to transform that into a sum of isometries multiplied by a constant. The same goes for singular value decomposition, $$Tv=s_1 \langle v,e_1 \rangle f_1+...+s_n\langle v,e_n \rangle f_n$$ where $e_1,...,e_n$ and $f_1,...,f_n$ are orthonormal bases of $V$, $s_1,...,s_n$ are singular values of $T$, and $v\in V$. Not sure where to go from there.

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, start with the polar decomposition $T = G P$ where $G$ is unitary and $P = \sqrt{T^*T}$ is positive semidefinite. Let $P = p R$ where $p>0$ is large enough that $I-R^2$ is positive semidefinite. Then $R = (R + i \sqrt{I-R^2})/2 + (R - i \sqrt{I-R^2})/2$, where $R \pm i \sqrt{I-R^2}$ are unitary.

$\endgroup$
3
  • $\begingroup$ Thanks for the reply. I don't see why $P$ is necessarily of the form $P=pR$? Also, why are $R\pm i \sqrt{I-R^2}$ unitary? I'm pretty confused to be honest. $\endgroup$ Commented Aug 10, 2016 at 19:28
  • $\begingroup$ $p$ is a scalar. Just divide $P$ by $p$. If $U = R + i \sqrt{I-R^2}$, $U^* = R - i \sqrt{I-R^2}$ and $U U^* = I$. $\endgroup$ Commented Aug 10, 2016 at 20:59
  • $\begingroup$ and what about the constant $a$ in $a(S_1+S_2)$? Thanks again. $\endgroup$ Commented Aug 11, 2016 at 1:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .