When is the intersection of two element-generated hereditary C$^{*}$-algebras non-zero Let $A$ be a unital C$^{*}$-algebra. Let $b$ and $c$ be non-zero positive elements in $A$. Let's suppose that I can find a unitary $v$ in $A$ such that $bvc\not= 0$ and, hence, such that $z:=bvcv^{*}\not=0$. 

Is this enough to prove that the intersection of the hereditary C$^{*}$-algebras $H_{1}=\overline{bAb}$ and $H_{2}=\overline{vcv^{*}Avcv^{*}}$ is non-zero? 

I realize that $z^{*}z\in H_{2}$ and that $zz^{*}\in H_{1}$, but I don't think this is enough. 

It's not enough. Take $A=M_2(\mathbb C)$, and 
$$
b=c=\begin{bmatrix} 1&0\\0&0\end{bmatrix}, \ v=\tfrac1{\sqrt2}\,\begin{bmatrix} 1&1\\1&-1\end{bmatrix}.
$$
We have $q:=vcv^*= \tfrac12\,\begin{bmatrix} 1&1\\1&1\end{bmatrix}$.
Then 
$$
bvc=\tfrac1{\sqrt2}\,\begin{bmatrix} 1&0\\0&0\end{bmatrix}\ne0,
$$
while 
$$
bAb=\mathbb C\,b,\ \ qAq=\mathbb C\,q,
$$
and the intersection of the two algebras is $\{0\}$.
