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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.

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  • $\begingroup$ And you need to prove that $A*A$ is hermitian (self-adjoint). $\endgroup$ – Valerin Nov 21 '13 at 21:11
  • $\begingroup$ @LuisValerin I don't know what that means... $\endgroup$ – Alti Nov 21 '13 at 21:15
  • $\begingroup$ Look as hint considerer to prove the following: Compute the product $A*A$ and prove that the principal minors(determinants of all orders) are positive. $\endgroup$ – Valerin Nov 21 '13 at 21:16
  • $\begingroup$ OK. Give me a minute to write the proof. $\endgroup$ – Valerin Nov 21 '13 at 21:17
  • $\begingroup$ $\langle A^* A h,h \rangle = \langle Ah,Ah \rangle = \Vert A h \Vert ^2$ $\endgroup$ – roger Nov 21 '13 at 21:23
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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

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  • $\begingroup$ Thank you. How did you know $<h,A^*Ah>=<Ah,Ah>$? $\endgroup$ – Alti Nov 21 '13 at 21:37
  • $\begingroup$ @ Alti: Thanks for the thanks! As for your question, for any matrix $B$ and vector $h$, $\langle h, B^*h \rangle = \langle Bh, h \rangle$. To see this, write everything out in terms of coordinates. If you need more info, post another connent and I'l explain further. For now, gotta go; work beckons! $\endgroup$ – Robert Lewis Nov 21 '13 at 21:44
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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.

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    $\begingroup$ you can raise the "$*$" to a superscript in Latex with the "^" notation; thus "A^*" yields $A^*$; works for other symbols as well: $A^2$, $A^3$, $A^{B^{C^D}}$, you name it! Incidentally, if you hover the pointer over rendered Latex and right click, you can open a window which shows the raw Latex; a great way to learn the stuff! $\endgroup$ – Robert Lewis Nov 22 '13 at 5:32
  • $\begingroup$ Luis Valerin- Thank you for providing a different method in approaching the problem! $\endgroup$ – Alti Nov 23 '13 at 4:31

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