It is known that a $n^2 \times n^2$ positive semidefinite matrix $A$ cannot always be written as a finite sum
$$
A=\sum\_{j} B\_j \otimes C\_j
$$
with $B\_j$ and $C\_j$ positive semidefinite matrices (of size $n \times n$). For example, it can be seen that the matrix
$$
\begin{pmatrix}
1 & 0 & 0 & 1 \\\
0 & 0 & 0 & 0 \\\
0 & 0 & 0 & 0 \\\
1 & 0 & 0 & 1 
\end{pmatrix}
$$
is not the finite sum of Kronecker products of positive semidefinite $2 \times 2$ matrices. 


Is the statement true if $A$ is positive definite? (i.e., $A$ is invertible?).


Edit: this question is a slight variant of a [previous question](https://mathoverflow.net/questions/43138/positive-elements-in-tensor-products).



Decomposition of positive definite matrices.
**The following is just a minor variation of [Martin Argerami's proof of the old question](https://mathoverflow.net/questions/43138/positive-elements-in-tensor-products/43198#43198).** I am even copying his equations and some of his text. If you are +1ing this post, please also +1 his one (if not already done).


Here is a counterexample for $n=2$. Let $\varepsilon $ be a positive real $<
\dfrac{1}{2}$. The matrix


$
a= \begin{bmatrix}
1 & 0 & 0 & 1-\varepsilon \\ 
0 & \varepsilon & 0 & 0 \\ 
0 & 0 & \varepsilon & 0 \\ 
1-\varepsilon & 0 & 0 & 1
\end{bmatrix}
\in \mathrm{M}\_{4}\left( \mathbb{C}\right)
$


is positive-definite, but it cannot be written as a sum of tensor products
of nonnegative-semidefinite $2\times 2$-matrices. Here is why:


Assume the contrary. Thus, $a$ is written in the form


$a=\sum\_{j} \left[ \begin{matrix}
\alpha \_{j} & \overline{\gamma \_{j}} \\ 
\gamma \_{j} & \beta \_{j}
\end{matrix} \right]
\otimes \left[
\begin{matrix}
\alpha \_{j}^{\prime } & \overline{\gamma \_{j}^{\prime }} \\ 
\gamma \_{j}^{\prime } & \beta \_{j}^{\prime }
\end{matrix} \right]
=\left[
\begin{matrix}
\sum\_{j}\alpha \_{j}^{\prime }\alpha \_{j} & \sum\_{j}\alpha \_{j}^{\prime }
\overline{\gamma \_{j}} & \sum\_{j}\overline{\gamma \_{j}^{\prime }}\alpha \_{j}
& \sum\_{j}\overline{\gamma \_{j}^{\prime }}\gamma \_{j} \\ 
\sum\_{j}\alpha \_{j}^{\prime }\gamma \_{j} & \sum\_{j}\alpha \_{j}^{\prime
}\beta \_{j} & \ast & \ast \\ 
\ast & \ast & \sum\_{j}\beta \_{j}^{\prime }\alpha \_{j} & \ast \\ 
\ast & \ast & \ast & \ast
\end{matrix}\right] $,


where $j$ ranges from $1$ to some positive integer $N$. Since each $
\begin{bmatrix}
\alpha \_{j} & \overline{\gamma \_{j}} \\ 
\gamma \_{j} & \beta \_{j}
\end{bmatrix}
$ is nonnegative-semidefinite, we have $\alpha \_{j}\geq 0$, $\beta \_{j}\geq 0$,
and $\alpha \_{j}\beta \_{j}\geq \left\vert \gamma \_{j}\right\vert ^{2}$ for
all $j$. Similarly, $\alpha \_{j}^{\prime }\geq 0$, $\beta \_{j}^{\prime }\geq
0$, and $\alpha \_{j}^{\prime }\beta \_{j}^{\prime }\geq \left\vert \gamma
\_{j}^{\prime }\right\vert ^{2}$ for all $j$.


Now, comparing the entries of $a$ in this equation, we get $\varepsilon
=\sum\_{j}\alpha \_{j}^{\prime }\beta \_{j}$ (from the $\left( 2,2\right) $-th
entry) and $\varepsilon =\sum\_{j}\beta \_{j}^{\prime }\alpha \_{j}$ (from the $
\left( 3,3\right) $-th entry). Taking the arithmetic mean of these two
equations, we get


$\varepsilon =\dfrac{1}{2}\left( \sum\_{j}\alpha
\_{j}^{\prime }\beta \_{j}+\sum\_{j}\beta \_{j}^{\prime }\alpha \_{j}\right)
=\sum\_{j}\dfrac{\alpha \_{j}^{\prime }\beta \_{j}+\beta \_{j}^{\prime }\alpha
\_{j}}{2}\geq \sum\_{j}\sqrt{\alpha \_{j}^{\prime }\beta \_{j}\beta \_{j}^{\prime
}\alpha \_{j}}$


(by AM-GM, since we are dealing with nonnegative reals). But
for every $j$, we have


$\sqrt{\alpha \_{j}^{\prime }\beta \_{j}\beta
\_{j}^{\prime }\alpha \_{j}}=\sqrt{\alpha \_{j}\beta \_{j}}\sqrt{\alpha
\_{j}^{\prime }\beta \_{j}^{\prime }}\geq \left\vert \gamma \_{j}\right\vert
\left\vert \gamma \_{j}^{\prime }\right\vert $


(since $\alpha \_{j}\beta\_{j}\geq \left\vert \gamma \_{j}\right\vert ^{2}$ and $\alpha \_{j}^{\prime }\beta \_{j}^{\prime }\geq \left\vert \gamma \_{j}^{\prime }\right\vert ^{2}$
), so this becomes


$\varepsilon \geq \sum\_{j}\underbrace{\left\vert \gamma
\_{j}\right\vert \left\vert \gamma \_{j}^{\prime }\right\vert }\_{=\left\vert 
\overline{\gamma \_{j}^{\prime }}\gamma \_{j}\right\vert } =
\sum\_{j}\left\vert \overline{\gamma \_{j}^{\prime }}\gamma \_{j}\right\vert
\geq \left\vert \sum\_{j}\overline{\gamma \_{j}^{\prime }}\gamma
\_{j}\right\vert $


(by the triangle inequality). But since $1-\varepsilon
=\sum\_{j}\overline{\gamma \_{j}^{\prime }}\gamma \_{j}$ (by comparing the $
\left( 1,4\right) $-th entry of the matrices in the above equation), this
becomes $\varepsilon \geq \left\vert 1-\varepsilon \right\vert $, what
contradicts the definition of $\varepsilon $.


