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The following Hermitian matrix is given for all $m \in \mathbb{Z}$:

$A_m = \left( \begin{array}{cc} m & i \\ -i & m\\ \end{array} \right)$

(i) Show that $A_1$ is not positive definite.

(ii) Show that $A_2$ is positive definite.

(iii) Determine all $m \in \mathbb{Z}$ such that $A_m$ is positive definite.

So my understanding is: when given a matrix, to check if it is positive definite, one calculates: $\overline{z}^{t}Az$. Would this always be the same as $z^{t}A\overline{z}$? (When I calculated it out, the answer seemed to be "no"). I got the former off wikipedia, but I think I've seen the later as well...

Using the first version, I get $\overline{z}^{t} A_{m}z = \left( \begin{array}{cc} m \overline{z_{1}}-i\overline{z_{2}}, & i\overline{z_{1}}+m\overline{z_{2}} \end{array} \right) \left(\begin{array}{c}z_{1} \\ z_{2} \end{array} \right) = mz_{1}\overline{z_{1}}-iz_{1}\overline{z_{2}}+i\overline{z_{1}}z_{2}+mz_{2}\overline{z_{2}}$ and since I'm examining whether they are positive definite, I think I should try to express as many terms as squared so I think I can write $=m|z_{1}|^{2}-iz_{1}\overline{z_{2}}+i\overline{z_{1}}z_{2}+m|z_{2}|^{2}$. Here is where I don't know what to do with regard to parts (ii) and (iii)... Is there some sort of way to factor or rearrange the terms to make a case for which $m\in \mathbb{Z}$ this expression must be $\geq 0$? I think I've managed part (i) since $m=1, 0 \neq z = (-1,-i) \Rightarrow 1-i(-1)(i)+i(-1)(-i)+1 = 1+i^{2}+i^{2}+1 = 0$. The problem is that I just got that through playing around and don't really know why it should be so...

Also, just to make sure, the same approach will work for a Hermitian matrix of higher order as well, right?

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    @J.M.: sorry, but I still can't understand your idea... First, I thought I had already worked out the $m=1$ case, right? Second, I don't see where, how, or why to make $a,b,c,d$ evaluate to something manifestly nonpositive... Also when you say quadratic form, you are referring to a specific type of mapping, or just any polynomial with terms of degree 2?2011-04-26

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The trace of you matrix is 2m and its determinant is m^2-1 thus you just need you m to be 1)strictly positive 2)m^2>1 thus Am is definite positive for any m integer greater than or equal 2

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In this case, you can compute the eigen values and find out for what $m$'s the eigen values are positive. (Caution: In general, if the matrix is positive definite the eigen values are positive. However it is not always the case that if all the eigen values are positive the matrix is positive definite. You need the geometric multiplicity of the eigen values should be equal to the algebraic multiplicity to conclude the matrix is positive definite (which happens to be true this case for m >1))

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    Sorry, the "transpose" has to be changed by "hermitian conjugate" in my comment.2011-04-26
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You could try to prove (and use) the following criterion: $ \begin{pmatrix} a & b \\ \bar{b} & c \end{pmatrix} $ is positive definite iff $a > 0$ and $ac - |b|^2 > 0$ (here $a,c$ are real and $b$ is complex).

Hint: apply to the vector $\begin{pmatrix} b \\ -a \end{pmatrix}$

This criterion can be generalized to higher dimensions: if $(a_{i,j})_{1 \leq i,j \leq n}$ is hermitian, it is positive definite iff for any $1 \leq k \leq n$, $\det \left( (a_{i,j})_{1 \leq i,j \leq k} \right) > 0$. To prove it, you need to show that the maximal subspaces on which a hermitian form is positive definite have the same dimension, and proceed by induction on $n$.

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    A related generalization: if $M$ is the block matrix \begin{pmatrix} A & B \\ -B & A \end{pmatrix}, since the block matrix U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & -iI \\ I & iI \end{pmatrix} is unitary and U M U^* = \begin{pmatrix} A + iB & 0 \\ 0 & A - iB \end{pmatrix}, you see that $M$ is positive definite if and only if $A+iB$ and $A-iB$ both are. (When $A = m$ and $B = i$ are $1 \times 1$ scalar matrices the condition is that $m \pm 1$ must both be positive, ie, that m > 1. This is nonelementary enough that I'm not seriously trying to help the OP here; just having some fun.)2011-04-26