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 followsScreenshot of the John Watrous TQI-notes

  • 2
    $\begingroup$ 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)? $\endgroup$ Commented Jun 6 at 20:40

1 Answer 1


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


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