# Clarification about the Alberti's Theorem proof given by Watrous in his condensed lecture notes

In the John Watrous condensed TQI lecture notes, an alternative proof of the Alberti's Theorem is given. He use an auxiliary lemma that states;

Lemma 4.9. Let $$P \in Pos(X)$$. It holds that $${inf}_{R\in PD(x)} \langle R,P\rangle\langle R^{-1},P\rangle = (Tr(P))^2$$.

Where $$Pos(x)$$ means positive semidefinite and $$PD(x)$$ means positive definite.

He begins the proof stating that since $$R = I$$(identity) is positive definite, then is clear that $$\langle R,P]\rangle\langle R^{-1},P\rangle \leq (Tr(P))^2$$. Then, for the opposite, '$$\geq$$', he states that, given $$\alpha$$ and $$\beta$$ real numbers $$\alpha^2 + \beta^2 \geq 2\alpha\beta$$ and therefore $$\alpha\beta^{-1} + \beta\alpha^{-1} \geq 2$$, assuming a spectral decomposition for $$R$$ and making use of the Hilbert-Schmidt product. $$R = \sum_{1}^{n} \lambda_i u_i u_{i}^{*}$$

$$\langle R,P\rangle\langle R^{-1},P\rangle = \sum_{i,j = 1}^{n} \lambda_i \lambda^{-1}_{j} (u_i^{*} P u_i)(u_j^{ *} P u_j) = \sum_i^{n} (u_i^{ *} P u_i)^{2} + \sum_i^{n}\sum_j^{n} (\lambda_i\lambda_j^{-1} + \lambda_j\lambda_i^{-1}) (u_i^{ *} P u_i)(u_j^{ *} P u_j)$$

In the last equation, in the last equality, i couldn't figure out what manipulation he used to achieve such equation. I've tried to find it for some time and i didn't reach anywhere close. How can he go from this $$\sum_{i,j = 1}^{n} \lambda_i \lambda^{-1}_{j} (u_i^{*} P u_i)(u_j^{ *} P u_j)$$ to this $$\sum_i^{n} (u_i^{ *} P u_i)^{2} + \sum_i^{n}\sum_j^{n} (\lambda_i\lambda_j^{-1} + \lambda_j\lambda_i^{-1}) (u_i^{ *} P u_i)(u_j^{ *} P u_j)$$

A print of his notes follows

• Isn't it just grouping pairs (i, j)? Maybe a factor of 1/2 is missing, or the double sum is only over different tuples (i, j) (e.g. with j>i)? Commented Jun 6 at 20:40

Start from $$\sum_{j=1}^n\sum_{i=1}^n\lambda_i\lambda_j^{-1}(u_i^\star PU_i)(u_j^\star Pu_j)$$ and split up the sum over $$i$$ into 3 terms: $$i, $$i=j$$ and $$i>j$$. $$\sum_{j=1}^n\sum_{ij}\lambda_i\lambda_j^{-1}(u_i^\star PU_i)(u_j^\star Pu_j)$$ Now, for every pair $$(i,j)$$ for which $$i>j$$, we can equally think of this as a pair $$(j,i)$$ for which $$j. So, this whole thing simplifies to $$\sum_{j=1}^n\sum_{i $$i$$ and $$j$$ are arbitrary indices, so let's just swap them in the third term. $$\sum_{j=1}^n\sum_{i Now the summation for the first and last terms is the same, so we might as well group them. $$\sum_{j=1}^n(u_j^\star PU_j)^2+\sum_{j=1}^n\sum_{i This is what you were after.