minimal projections in matrix-algebras Consider $A=\{ \begin{pmatrix} T & 0 \\ 0 & T \end{pmatrix}: T\in M_2(\mathbb{C})\}\subseteq M_4(\mathbb{C})$ and $p= \begin{pmatrix} 1 & 0&0&0 \\ 0 & 0&0&0\\0 & 0&1&0\\ 0 & 0&0&0 \end{pmatrix}\in A.$
We defined $p$ to be a projection, if $p^2=p=p^*$. I have already shown that $p$ is a projection.
$p$ is a minimal projection in $A$, if: for all projections $0\neq q\in A$ such that $q\le p$, $\Rightarrow \; p=q$. 
$q\le p$ means, that $p-q$ is a positive operator. Equivalent is: $image(q)\subseteq image(p)$. 
The claim is: $p$ is minimal in $A$, but $p$ isn't minimal in $M_4(\mathbb{C})$. 
-My solution for p minimal in A: It is $M_2(\mathbb{C})\cong A$ (Consider the linear, multiplicative, adjoint-preserving bijective map $\gamma:M_2(\mathbb{C})\to A, T\mapsto \begin{pmatrix} T & 0 \\ 0 & T \end{pmatrix}$). Therefore, T is minimal in $M_2(\mathbb{C}) \iff \begin{pmatrix} T & 0 \\ 0 & T \end{pmatrix}$ is minimal in $A$. Consider $T=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},$ it is $\dim (image(T))=1$ and $T^2=T^*=T$, i.e. $T$ is a minimal projection in $M_2(\mathbb{C})$. It follows, that $\gamma(T)=p\in A$ is a minimal projection in $A$.
But why isn't $p$ minimal in $M_4(\mathbb{C})$?
Edit: Sorry, i was too fast. $p$ isn't minimal in $M_4(\mathbb{C})$, because consider $q= \begin{pmatrix} 1 & 0&0&0 \\ 0 & 0&0&0\\0 & 0&0&0\\ 0 & 0&0&0 \end{pmatrix}$. It is $q=q^*=q^2$ and $0\le q\le p$, but it isn't $p=q$.

Here the answer again: $p$ isn't minimal in $M_4(\mathbb{C})$, because consider $q= \begin{pmatrix} 1 & 0&0&0 \\ 0 & 0&0&0\\0 & 0&0&0\\ 0 & 0&0&0 \end{pmatrix}$. It is $q=q^*=q^2$ and $0\le q\le p$, but it isn't $p=q$, because p has a smaller image than p.
