# Show that if $\sum_{j=1}^m E_j=I$ and $E_j \geq 0$ then the dual basis cannot be positive semi-definite.

Consider a $d$-dimensional Hilbert space $\mathcal H$, and the real vector space of Hermitian operators $H_d$. Define a set of Hermitian matrices $$\beta := \{E_1,E_2,...,E_m \}, \qquad m \geq d^2$$ which are positive semi-definite, $E_j \geq 0$, and satisfy $$\sum_{j=1}^m E_j=I,$$ the identity matrix. Furthermore, the set $\beta$ is a frame, meaning that $\text{span}_{\mathbb R} \beta = H_d$. Define the dual frame to $\beta$ $$\tilde \beta := \{ Q_1,Q_2,...,Q_m\}$$ which is a collection of Hermitian matrices that satisfy $$\text{tr}(E_i Q_j) = \delta_{ij}.$$ Any positive semi-definite matrix $\rho$ can then be written in terms of the dual frame as $$\rho = \sum_{j=1}^m a_j Q_j$$ where $a_j = \text{tr}(\rho E_j) \geq 0$. From these properties I am trying to show that in order for the above properties to be true, at least one of the $Q_j$ matrices must not be positive semi-definite. Can anybody show me where to start?

Note that the statement is not true when $d=1$.

Also, by the Kronecker delta condition, the $E_i$s must be linearly independent, for, if $\sum_i c_iE_i=0$, then $0=\operatorname{tr}\left(\sum_i c_iE_iQ_j\right)=\sum_i c_i\operatorname{tr}\left(E_iQ_j\right)=c_j$ for each $j$. Moreover, as the $E_i$s span the real vector space $H_d$ of all Hermitian matrices, they must form a basis of $H_d$. It follows that $m$ must be equal to $d^2$.

So, let us assume that $m=d^2>1$. Suppose the contrary that every $Q_j$ is positive semidefinite. Obviously, by the Kronecker delta condition, all $Q_j$s are nonzero.

Let $v\ne0$. By assumption, $vv^\ast$ can be written as a nonnegatively weighted sum $\sum_ja_jQ_j$. Since each $Q_j$ is positive semidefinite, $a_j>0$ if and only if $Q_j$ is a positive multiple of $vv^\ast$, and $a_j=0$ otherwise.

In other words, for each $v\ne0$, there exists at least one $Q_j$ that is a positive multiple of $vv^\ast$. But this is a contradiction, because when $d>1$, there exist an infinite set of pairwise linearly independent vectors $v$.

• Thanks for the response, I'm having a hard time understanding why this is a contradiction, are you able to elaborate? Why is this not a contradiction when we allow some $Q_j$ to not be positive semi-definite? – gene Mar 7 '18 at 16:19
• @gene It is sometimes possible to write a rank-1 PSD matrix $vv^T$ as a positively weighted sum of Hermitian matrices of rank $>1$. So, if some $Q_j$ is not PSD, we cannot relate each $Q_j$ to a unique direction $v$ in our proof. – user1551 Mar 7 '18 at 16:36
• E.g. suppose \begin{align} & E_1=\frac16\pmatrix{2&1\\ 1&1}, \ E_2=\frac16\pmatrix{2&-1\\ -1&1}, \ E_3=\frac16\pmatrix{1&i\\ -i&2}, \ E_4=E_3^\top,\\ & Q_1=\frac12\pmatrix{4&3\\ 3&-2}, \ Q_2=\frac12\pmatrix{4&-3\\ -3&-2}, \ Q_3=\frac12\pmatrix{-2&3i\\ -3i&4}, \ Q_4=E_3^\top. \end{align} Then $E_1=\frac14(Q_1+Q_2+Q_3+Q_4)$. Each weight $a_j\,(=\frac14)$ here is positive, but none of the $Q_j$s has rank one. – user1551 Mar 7 '18 at 22:35