Prove positive, semi-definite Let $A \in M_2(\mathbb C)$ and let $A^*$ denote the conjugate transpose of $A$. I need to show that $A^{*}A$ is positive semi-definite. 
So I need to show $$\quad\quad \langle  A^{*}\!A\,h,\;h\rangle \geq 0,\mbox{ for all }h \in M_2(\mathbb C)$$. 
I don't know how to approach this, any help is appreciated. 
 A: I think you want $h \in \Bbb C^2$, not $h \in M_2(\Bbb C)$.
That being said:
To see that $A^*A$ is postive semi-definite, simply note that $\langle h, A^*Ah \rangle = \langle Ah, Ah \rangle$ is real, since for any $z$, $\langle z, z \rangle = \langle z, z \rangle^*$.  Also, $\langle Ah, Ah \rangle \ge 0$, since we always have $\langle z, z \rangle \ge 0$.  This shows $A^*A$ is positive semi-definite.  QED
Nota Bene:  You don't need to directly show, as Luis Valerin's comment suggests, that $A^*A$ is Hermitian, but that is easy in any event: $(A^*A)^* = A^*A^{**} = A^*A$.  End of Note.
Hope this helps.  Cheerio, 
and as always,
Fiat Lux
A: To prove that $A*A$ is definite positive you need to prove two things. 
First, that is self-adjoint, i.e. $(A*A)=(A*A)*$. 
And second, you have to prove that  $A*A$ is positive. For this part there exist many forms to approach. The most easy is to prove that the minors of this matrix are positive. See http://en.wikipedia.org/wiki/Minor_(linear_algebra). 
The first part follows easily because, $(A*A)*=(A*)((A*)*)=A*A$. For the second part, consider 
$$A=\left(\begin{array}{rcl}
a & b\\
c & d \end{array} \right)$$ 
then $$A*=\left(\begin{array}{rcl}
\overline{a} & \overline{c}\\
\overline{b} & \overline{d} \end{array} \right) $$ 
so $$A*A=\left(\begin{array}{rcl}
\overline{a} & \overline{c}\\
\overline{b} & \overline{d} \end{array} \right) \left(\begin{array}{rcl}
a & b\\
c & d \end{array} \right)=\left(\begin{array}{rcl}
\overline{a}a+\overline{c}c & \overline{a}b+\overline{c}d\\
\overline{b}a+\overline{d}c & \overline{d}d+\overline{b}b \end{array} \right) $$
$$\left(\begin{array}{rcl}
a^2+c^2 & \overline{a}b+\overline{c}d\\
\overline{b}a+\overline{d}c & d^2+b^2 \end{array} \right) $$
clearly the determinant of order 1 is $a^2+c^2\geq 0$ and the second order is also positive.
